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The Cambridge Technical Series 
General Editor: P. Abbott, B.A. 



AN INTRODUCTION TO 

APPLIED MECHANICS 



BY THE SAME AUTHOR 

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[In the Broadway Series of Engineering Handbooks.] 

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AN INTRODUCTION 

TO 

APPLIED MECHANICS 



BY 



EWART S. ANDREWS, B.Sc. Eng. (Lond.) 

Lecturer in the Engineering Department of the Goldsmiths* College, 

New Cross, and of the Westminster Technical Institute 

Formerly demonstrator and lecturer to the Mechanical Engineering 

Department of University College, London 

Member of the Council of the Concrete Institute 



Cambridge : 

at the University Press 
1920 






CAMBRIDGE UNIVERSITY PRESS 

C. F. CLAY, Manager 

LONDON : Fetter Lane, E.G. 4 

m 




NEW YORK : G. P. PUTNAM'S SONS 

BOMBAY, CALCUTTA, MADRAS: MACMILLAN AND CO., Ltd. 

TORONTO: J. M. DENT AND SONS, Ltd. 

TOKYO: THE MARUZEN-KABUSHIKI-KAISHA 






First Edition 1915 
Reprinted 1920 



All rights reserved 



%l- /^,^6/ 



PREFACE 

"lYyTANY engineering and architectural teachers have found 
-^^-*- that appUed mechanics is not an easy subject to teach, 
and most students have discovered that it is a difficult subject 
to understand. In searching for the reason for this unfortunate 
state of affairs, the author came to the conclusion that the 
treatment of the older form of text-book was too much that of 
applied mathematics — a kind of exercise-ground for algebraic 
manipulation — and that many of the more modern books that 
have attempted to remedy this weakness have given too much 
engineering apphcation of the principles of mechanics without 
sufficient explanation of those principles. 

The aim of the present book is to present the elementary 
principles of mechanics in accurate though clear terms and to 
show the application of those principles to the simpler problems 
arising in engineering and architectural appUcations. The general 
treatment is based more upon graphical conceptions than upon 
purely mathematical analysis because experience shows that 
the mind of the engineering student reasons more clearly from 
diagrams than from symbols. 

A number of simple experiments have been given, principally 
those which require the simplest form of apparatus. It is not 
suggested that the experiments given are all that are desirable 
in a laboratory course, but it is believed that sufficient have 
been given to make the principles clear. It may be pointed out 
here that there is some danger in attempting to learn principles 
merely by experiments with simple (and usually inaccurate) 



vi PREFACE 

apparatus. Before the student can hope to obtain valuable 
results from experiments, he must learn to make accurate 
readings of his instruments and to make corrections for the 
errors that may arise. Some authorities seem to suggest that 
experiment is of much greater importance to engineers than 
reasoning, but it should be borne in mind that training is required 
for good experimental work as well as for anything else, and in 
the author's opinion many engineering students who attempt 
to gather a knowledge of mechanical principles from experiment 
have not had sufficient preliminary training in experimental 
method. If our reasoning is based upon experimental laws 
and not upon dogmatic mathematical conceptions we shall 
probably make greater progress in elementary work by using 
experiment as an illustration of the results of our reasoning 
than by attempting to deduce the principles from the results 
of our experiments. 

The great value of training in experimental work — and 
thorough training is essential — lies in the direction of research 
work which comes when we have understood the principles 
based upon the earHer researches of others. 

It is hoped that this book will be found of value as a class- 
book in the junior classes of Engineering Colleges and in Public 
Schools that have an engineering side. 

The author wishes to express his gratitude to Mr J. B. Peace, 
M.A., of Emmanuel College, Cambridge, for much valuable 
criticism and assistance with the proofs, and to the publishers 
for the great help that they have given in the preparation of 
the diagrams. 

E. S. A. 

Goldsmiths' College, 

New Cross, S.E. 

May 1915. 



CONTENTS 

CHAPTER I 

FORCES AND OTHER VECTOR QUANTITIES 

Diagrammatic representation of forces — Resultant of a system of forces ; 
triangle of forces — Resolution of forces — Equilibrium; equilibrant — 
Vector polygon construction — ^Experimental errors . Pages 1-15 

CHAPTER II 
MOMENTS AND LEVERAGE 

Positive and negative moments — ^The principle of moments — Reactions 
on a beam — Stability of a wall — Lever safety valve — Equilibrium of 
a body under three forces — Link and vector polygon construction — 
Couples lS-35 

CHAPTER III 

WORK, POWER AND ENERGY 

Definitions — ^Kinetic and potential energy — Conservation of energy — 
Useful energy — Work done by a variable force — Work against 
resistance — Graphical representation of effort and resistance — Mean 
effort 36-51 

CHAPTER IV 
MACHINES AND EFFICIENCY 

Wheel and axle and crow-bar — Mechanical advantage; efficiency of 
machines ; velocity ratio — ^The inclined plane — ^The screw and screw- 
jack — Reversing machines — Pulley tackle — Weston's pulley block — 
Actual performance of machines — Indicated and brake horse- 
power 52-82 

CHAPTER V 
VELOCITY AND ACCELERATION 

Uniform velocity — Velocity variable in magnitude — Velocity and space 
curves — Acceleration — Relation between acceleration, velocity and 
space curves — Equations of motion for constant acceleration — 
Gravity acceleration — Limits of use of simple formulae — Distance 
moved in a particular second ..... 83-102 



viii CONTENTS 

CHAPTER VI 

VELOCITY CHANGE IN DIRECTION; RELATIVE VELOCITY 

Combination of velocities — Change of velocity — Relative velocity 

103-112 

CHAPTER VII 

KINETIC ENERGY AND MOMENTUM 

Measurement of kinetic energy — Connection between force and accelera- 
tion — ^Momentum — Importance of acceleration in traction problems 

113-123 

CHAPTER Vin 

NEWTON'S LAWS OF MOTION; IMPACT 

Newton's Laws — Impact and impulse — Equality of momentum before 
and after impact — Recoil of guns — Pile-drivers . . 124-138 

CHAPTER IX 

STRESS AND STRAIN 

Definitions — Hooke's Law — Stress-strain diagrams for mild steel, cast 
iron and concrete — Elastic moduli — Factor of safety — Resilience — 
Stress due to sudden loading — Temperature stresses . 139-160 

CHAPTER X 

RWETED JOINTS; THIN CYLINDERS 

/ Forms of rivet heads and joints, and diameter of rivets — Methods in which 
a joint may fail — Efficiency of joint — Strength of thin cylinders 
and pipes 161-173 



^ 



CHAPTER XI 

THE FORCES IN FRAIVIED STRUCTURES 

Kinds of framed structures — Relation between bars and nodes in a perfect 
frame — Curved members — Reciprocal figure construction — Distinction 
between ties and struts — The method of moments . . 174-187 



CHAPTER XII 

BEAMS AND GIRDERS 

Shearing force and bending moment — Diagrams for standard cases of 
loading for cantilevers and simply supported beams — Graphical 
construction 188-202 



CONTENTS ix 

CHAPTER XIII 
CENTRE OF GRAVITY AND CENTROID 

Centre of gravity by moments — Centre of gravity as balance point and by 
inspection — Centroid of an area — Centroid of triangle, quadrilateral, 
trapezium, semicircle and parabola — Centre of gravity of pjnramids 
and cones — Graphical construction for centroid — Kinds of equilibrium 

203-223 

CHAPTER XIV 

FRICTION AND LUBRICATION 

Static and kinetic friction — Coefficient of friction and angle of friction — 
Rolling friction — Inclined plane and screw with friction — ^Angle of 
repose — ^Efficiency of a screw — ^Lubrication . . , 224-241 

CHAPTER XV 
MOTION IN A CURVED PATH 

Hodograph — ^Uniform motion in a circle — Centripetal and centrifugal 
force — Railway curves and motor tracks — Centrifugal governors — 
Balancing rotating parts — Projectiles .... 242-257 

CHAPTER XVI 
MECHANISMS 

Crank and connecting-rod mechanism — Instantaneous or virtual centre — 
Watt's parallel motion — Quick-return mechanism — Toggle mechanism 
— Cams and wipers — Pawl and ratchet mechanism . 258-272 

CHAPTER XVII 
BELT, CHAIN AND TOOTHED GEARING 

Belt gearing — Velocity ratio — Speed-cones — Sizes of cones for keeping 
belt taut — Belt reversing gear — ^Belt drive for inclined axes — Toothed 
gearing — Rack, spur, bevel, spiral and worm gearing — ^Toothed gear 
trains — Idle gear wheels — Back gear for lathes — Reversing drive for 
lathe lead screw — Bevel gear reversing train . . . 273-293 

APPENDIX 
Sum curve construction 294 

Trigonometrical relations 297 

Mathematical tables ........ 298 

Note. — ^The tables of Logarithms and Anti-Logarithms are taken from 
Tables and Constants to Four Figures by W. Hall; those of Natural 
Sines, Cosines and Tangents from Four-figure Tables by C. Godfrey 
and A. W. Siddons. 

Answers to Exercises 310 

Index 314 



f 



CHAPTER I 

FORCES AND OTHER VECTOR QUANTITIES 

Quantities which can be represented in magnitude and 
direction by straight lines are called vector quantities; the 
length of the straight line represents the magnitude of the 
quantity to some chosen scale and the direction of the straight 
line, as indicated by an arrow-head, represents the direction 
in which the quantity acts. The term vector is used in contra- 
distinction to the term scalar, scalar quantities being those 
which have magnitude only. Length is a good example of a 
scalar quantity ; time is another. When we say that a body is 
10 inches long we know everything about the length, but there 
are some quantities which are not fully defined until we know 
the direction as well. Forces and velocities are familiar examples 
of vector quantities ; a vertical force of 10 lbs. is not the same 
thing as a horizontal force of 10 lbs. It should, however, be 
remembered that the direction and magnitude of a force does 
not tell us all that we wish to know about it; we must also 
know the position of the force, i.e., the actual position of the 
line of action of the force, because a force may be regarded as 
acting at any point in its line of action. 

The scientific definition of a force is as follows: "^ force 
is that which alters^ or tends to alter, the motion of a body in a 
straight line.'' This definition is based upon Newton's first 
law of motion which states that ** a body continues in a state 
of rest or uniform motion in a straight line unless it be acted 
upon by some external force." It is not, however, essential 
to understand fully this definition of force at the present stage, 
because the idea of a force as a push or a pull is quite clearly 
understood by most people. 

A. LI, 1 



FORCES AND OTHER 



[CH. 



Weight. It is one of the fundamental laws of nature that 
between all bodies there exists a force of attraction and the 
earth exerts upon all bodies a force called the force of gravity 
tending to pull the body towards the centre of the earth. The 
weight of a body is the force exerted by gravit}^ upon it and is 
the most familiar case of a force. 

Unit of force. The weight of a given quantity of matter is 
found to va,Ty with the latitude; it is about one-half per cent, 
greater at the poles than at the equator. We will take as our 
imit of force the weight of one pound in London; the pound 
being the quantity of matter in a standard cylinder of platinum 
preserved in London by the Board of Trade. This unit is often 
known as the "engineer's unit" or the "gravitation unit," to 
distinguish it from the "absolute unit " used by physicists. The 
absolute unit, which is the force required to produce a definite 
change of motion in a given mass in an assigned time, is inde- 
pendent of locality and is the more scientific of the two. As 
however engineers must be able to express their data and their 
results in the units in common use in their profession, and as 
the simultaneous use of two systems of units only leads to con- 
fusion, we shall confine ourselves to the engineer's unit as defined 
above. 

Diagrammatic representation of forces. Referring to Fig. 1, 
suppose that a force Fi acts through a point ^ in a body; as 








Force diagram 



Fig. 1. 



Vector diagram 



previously noted, it is better to speak of a force as acting through 
a point than as at a point. If now in some convenient position 



I] VECTOR QUANTITIES 3 

we draw a line 1, 2 parallel to Fi and of length to represent its i 
magnitude to some convenient scale, 1, 2 will be the vector for I 
the force F^, the arrow-head representing the direction in which j 
the vector is to be considered as acting. A very convenient 
method of indicating the force in many cases consists in number- 
ing or lettering the space on each side of the force in the force 
diagram. The force is then denoted by the numbers or letters 
between which it acts ; thus the force Fj in the figure is called 1, 2, 
so that spaces on the force diagram correspond to points on the 
vector diagram. This is often referred to as " Bow's notation." 

Resultant of a system of forces; triangle of forces. The 
resultant of a system of forces acting upon a body may be defined 
as the single force which will have the same effect upon the body 
as the combined effect of the separate forces. 

Suppose that a second force F^ acts through a point B in 
the body and let the lines of action of the two forces intersect 
at the point G. On the vector diagram, starting from the point 
2, set out a length 2, 3 equal in length to the force F2 to the scale 
already chosen for F^ and parallel to F2 in the direction of its 
arrow-head, and join 1, 3, the triangle 1, 2, 3 being commonly 
referred to as the triangle of forces. Then 1, 3 represents in | 
magnitude and direction (but not in position) the resultant R * 
of the two forces. It will act through the intersection C of the 
lines of action of the two forces, so that by drawing as shown 
in dotted lines a line R through C parallel to 1, 3 we can say 
that R is the resultant of the two forces F^ and F^ and that its 
magnitude is given by the length 1, 3. 

It should be noted that the direction of the resultant is from 
the first point on the vector diagram to the last, i.e., from 1 to 3 ; 
by keeping this fact always in mind we shall avoid the confusion 
that sometimes arises. Further the resultant does not act in the 
line 1, 3 but through the intersection C of the two forces. 

To make quite clear the plotting of the vector diagram 
suppose that F^ is 12*5 lbs. and i^'g is 8 lbs. and that the vector 
diagram is drawn to a scale 1" = 10 lbs., then 1, 2 will be drawn 
125 inches long and 2, 3 will be made "8 inch long and if I, 3 
is 1*7 inches long, the resultant R of the two forces will be equal 
to 17 X 10 = 17 lbs. and will be in the direction shown. We 
need not restrict ourselves to actually measuring the resultant 

1—2 



FORCES AND OTHER 



[CH. 



R; we may calculate it when required by means of the trigono- 
metrical methods of the solution of triangles. When we do 
calculate in this way it is not necessary to draw the triangle 
of forces accurately to scale. We do not propose to give a rigorous 
proof of this result at present but it may easily be verified experi- 
mentally in the following manner: 

Experiment i. Upon a suitable drawing board A (Fig. 2) arranged 
vertically, to which is fixed a piece of paper P, fix a spring balance by a string 
to a point B and connect a string to the hook of the balance and tie it to a 
ring D to which is tied another string connected to a weight F^. Tie a third 
piece of string to the ring and pass the other end of it over a freely-mounted 




Fig. 2. Experiment on Triangle of Forces. 

pulley C and connect a weight Fi to its end. After the strings have come 
to rest, trace their directions upon the paper and remove the latter; then 
at some convenient place at the side draw to a suitable scale 1, 2 parallel 
to the portion DC of the string and of length to represent the weight Fi, and 
draw 2, 3 vertically and make it represent the weight F^ to the same scale 
and join 1, 3. Then 1, 3 will be found to equal a length representing the 
reading upon the spring balance to the previously chosen scale and will be 
found parallel to BD. The pull in the portion CD of the string is equal to 
the weight Fi on the end of it if there is perfect freedom of movement of the 
bearings of the pulley, so that our experiment will have verified the law of 
the Triangle of Forces. (This equality of tension on the two sides of the 
pulley follows from the Principle of Moments, p, 18.) Now suppose that one 
of the weights J^i or F., is changed ; the effect of this change will be that the 
strings will move and will finally come to rest in a new position ; the directions 
and magnitudes of the forces will again be found to follow the triangle law. 



I] VECTOR QUANTITIES 6 

Notation to represent vector addition. We have seen that 
the resultant E is equivalent to the combined action of the 
two forces F^ and i^g. B is then said to be the vector sum of 
Fi and F,^. We may write this symbolically as 

R = Fi^ F^. 

The -H-, which is a modified plus sign, indicates that the 
addition is not a mere numerical or algebraic one, but that the 
directions of the forces F^y F2 are taken into consideration in 
effecting the addition. 

Numerical Example — Thrust on a steam-engine foundation. 
As a simple numerical example of the triangle of forces take 




V V/'/^^ ^/'^yy/y/j'/'^^y' •^/'^-'y'/^yyy^^^^ 



P*1500 



WS2000N' 




Q = 2500 



Fig. 3. Thrust on Steam -Engine Foundation. 

the horizontal steam-engine shown in Fig. 3. The force or 
thrust upon the foundations, at the main bearing, for the position 



6 FORCES AND OTHER [ch. 

\ 



. shown of the crank-pin C and connecting-rod i?, is made up of 
Hhe weight W of the flywheel, shaft, etc., and the pull P exerted 



je 



by the piston. Take W = 2000 lbs. and P = 1500 lbs. and set 
out ab horizontally to a convenient scale to represent P, say 

jl" = 1000 lbs. (1000 lbs. forms a very convenient unit for 
ngineering calculations and has been called a " kip ") ; then 

"set out be vertically to the same scale to represent 2000 lbs. 
and join ac. Then ac gives the resultant thrust upon the founda- 
tions, a being the first point and c the last. 

ac if scaled off will be found to be 2500 lbs. (25 kips), 
but we should note that we can find it by calculation in this 
case as easily as by scaling off, and consequently we need not draw 
accurately to scale, i.e., because abc is a right-angled triangle, 

ac^ = ab^ -f bc^, 
i.e. ac = s/ab^Tbc^ = ^15002 4-20002 = 2500 lbs. 

If we want to find the angle 6 by trigonometrical calculation 

we note that ^i 

tan^ = ^=ff^= 1-3333, 

and from trigonometrical tables we find that 9 = 48° 35' nearly. 
It should again be noted that we have the choice of actually 
drawing the vector figure to scale and solving the problem 
graphically, or of using trigonometry or other mathematical 
means of calculation. The student should endeavour to be able 
to use both methods, each being appropriate in certain cases. 
If, however, at this stage he has no knowledge of trigonometry, 
this need not act as an obstacle to him; he can always use the 
graphical solution. We wish, however, to point out that many 
students fall into the mistake of never learning the mathematical 
method at all, and consequently waste time in many problems. 
A brief explanation of trigonometry is given in the appendix. 
In this case therefore we should write 2500 = 1500 -tt- 2000. 

Resolution of forces. Let F, Fig. 4, represent a force acting 
through a point and let OX, OF be two lines passing through 
at any inclination whatever. Suppose that we wish to know 
the forces acting in these two directions which will have a resultant 
equal to F. 

Set out a length 1, 2 to represent the force F: from one end, 
say 1, draw 1, 3 parallel to OX and from the other end draw 



I] 



VECTOR QUANTITIES 



2, 3 parallel to OY , the intersection giving the point 3. Then 
1, 3 represents the force Fx which acting in the direction OX will 
combine with Fyt represented by 3, 2, acting in the direction 
Yy and have a resultant equal to F ; this follows from the rule, 
that we have already explained, that the resultant of any two ) 
forces is given in magnitude and direction by the third side of 
a triangle the other two sides of which represent the two forces I 
in magnitude and direction. The force J^ is then said to be 
resolved into the forces Fx and Fy> which are called the com- 
ponents of the force F in the two given directions. 




Force diagram 







Vector diagram 





Fig. 4. Resolution of Forces. 



In an exactly similar manner the force F could have been 
resolved into components Fz, Fjj in the directions OZ, OU, bs 
shown in dotted lines intersecting at the point 4. We will 
again emphasize the fact that it is not necessary to measure 
the lengths 1, 3 and 2, 3; we may calculate them by trigono- 
metrical or other methods whenever convenient. It is important 
to note that when components in two directions at right angles 
are considered, the ttvo components are called the resolved parts of 
the force in the two directions and that a force has no ^'resolved part" 
in a direction at right angles to its line of action. 

Numerical Example on resolution of forces. A barge 
A {Fig. 5) is being pulled along a canal by a horse which exerts a 
force of 150 lbs. in a direction at 20° to the centre line of the canal. 
Find the force urging the barge forward and that tending to pull 
it into the side. We require to find in this case the components 
of the force of 150 lbs. in the direction AC and at right angles 
to AC. 



8 



FORCES AND OTHER 



[OH. 



Draw a line ac parallel to AC, and drawing ab in a direction 
at 20° to oc to a scale say I'' = 50 lbs., i.e., making ab = 3'', draw 
be perpendicular to ac. Then by measurement we should have 





Fig. 5. 

be = rOS" = 51*5 lbs. = the force tending to pull the barge into 

the side and ac = 2-82^ = 141 -0 lbs. = the force urging the barge 

forward along the centre of the canal. By calculation we should 

have 

be = 150 sin 20° = 150 x -3420 = 51-3 lbs. 

ac = 150 cos 20° = 150 x -9397 = 141-0 lbs. 

Equilibrium ; equilibrant. If a body is at rest or is moving 
without altering its velocity, the forces acting upon it are said 
to be " in equilibrium." Ji such is the case, there is nothing 
tending to " alter its condition of rest or uniform motion " so 
that there is no resultant force acting upon it. We see therefore 
that the first essential of equilibrium of a body is that the resultant 




Force diagram 



Vector diao:ram 



Fi2. 1. 



of all the forces acting upon it shall be zero. This means that the 
first and last points of the vector diagram must coincide, because 
the distance between the first and last points gives us the value 
of the resultant. 



VECTOR QUANTITIES 



9 



Referring again to Fig. 1, we see that the body is not in 
equilibrium under the action of the forces F^, F^ because their 
resultant R is given by 1, 3 which is not zero. Now the single 
force that has to be added to a system of forces acting upon 
a body to bring the body into equilibrium is called the equilibrant. 
The equilibrant must be equal and opposite to the resultant, because 
the system may, as we have seen, be considered as replaced 
by the resultant and our forces then become reduced to two — 
the resultant and the equilibrant — and the resultant oi: combined 
effect of these two forces must be zero for equilibrium. The 
only way for two forces to reduce to zero is for them to be 
equal and opposite ; the only way for instance for the force E to 
be neutralised or equilibrated is for the equilibrant to be equal 
to 3, 1. 

Warning; forces acting round a triangle. Care should be 
taken to distinguish between the following cases: 





Force diagram 



Vector diagram 



Pig. 6. 



(a) Three forces acting at a point and having their vector 
figure a closed triangle. This is the case that we have already 
considered and is shown diagrammatically in Fig. 6. 

(6) Three forces acting on a body round the sides of a triangle 
are not in equilibrium even if the sides of the triangle are pro- 
portional to the forces. This is shown in Fig. 7. In this case 
the three forces do not meet at a point and the two forces F^ and 
F2 have a resultant acting through B which is equal to i^j -h- i^^ = ^3 
parallel to ^C (if the forces are proportional to the sides of the 
triangle). The three forces are therefore equivalent to two equal 



10 FORCES AND OTHER [ch. 

a.nd opposite parallel forces F^ at a distance x apart. These 
two forces will tend to turn the triangle round in the direction 




Fig. 7. 

of the arrow O and form what is called a couple (which we shall 
consider more fully later). 

More than two forces. Vector polygon construction. 
Up to the present we have dealt with two forces only, but the 
idea of vector addition is applicable to any number of forces. 
Take for instance four forces F^FJF^F^ (Fig. 8) : to some con- 
venient scale draw 1, 2 parallel to represent F^ in magnitude 
and direction; then starting from 2 draw 2, 3 to represent 
^2> then 3, 4 to represent J^^g; and finally 4, 5 to represent F^. 
Then the resultant B of the whole system of forces will be given 
in magnitude and direction by the line 1, 5 joining the first 
point of the vector figure to the last. 

This is the general case of vector addition and can be expressed 
in words as follows : The resultant or sum of a number of vector 
quantities (i.e., those having magnitude and direction, such as 
forces and velocities or speeds) is obtained by placing them end 
to end, preserving their directions and a continuous sense of their 
arrow-heads ; the final step from the beginning of the first vector 
to the end of the last is the resultant or vector sum. 

The reader may find this definition rather difiicult to follow 
at first, but if he reads it carefully in connection with the 
figure the meaning should become clear. To express this result 
generally by a formula, where there are altogether n forces 
(where n is a whole number), the first of which is F^ and the 
last F^i we should write 

R = F^^F^^,,..F^, 



VECTOR QUANTITIES 



11 



This formula should be regarded merely as a symbolic or 
shorthand way of expressing the statement in italics. 

Proof. This principle may be proved by repeating the con- 
struction for two forces, thus: It follows by the triangle of 




• Polygon 



\«12 



> Line of Pressure 





Fig. 8. Vector Polygon construction. 



forces that the length 1, 3 (shown in chain dotted lines) represents 
the resultant JSj^g of the forces F-^, F2I it must act through 
their intersection x as shown on the force figure; let the line 
of action of i?i ., intersect the third 
force Fq ia y. Now 1, 4 represents 



r 

the resultant of i?i2 ^^^ -^3 ^^^ 
therefore of jP^, F2 and F^; draw 
therefore through y a line parallel 
to 1, 4 to intersect F^ as shown. 
The force 1, 5 clearly represents the 
resultant of 1, 4 and F^, i.e., of F^, 
F2, Fq and F^, and gives therefore the 
resultant R required and by drawing g 
a line parallel to 1, 5 through the 
point last obtained we get the line 
of action of R. 

The line x, y, etc. is often called 
the line of ^pressure because one of its principal practical applica- 
tions arises in problems relating to walls, arches and similar 






Fig. 9. 



12 



FORCES AND OTHER 



[OH. 

this 



structures. We shall deal later with an extension of 
graphic al c ons true tion . 

Order of taking the forces. It should be noted carefully 
that this construction gives the same result no matter what 
be the order in which the forces are taken, although the order 
will make an alteration in the shape of the vector polygon and 
for general convenience it is best as a rule to take them in turn. 
Fig. 9 for instance shows the vector polygon for the forces taken 
in the order ^2> -^4» ^8» -^i- The magnitude and direction of 
the resultant R is the same as before. 




Fig. 10. Experiment on Polygon of Forces. 



Experiment 2. This principle may be veri6ed experimentally by an 
apparatus similar to that employed for the triangle of forces. Connect two 
additional pulleys E, (Fig. 10) to the apparatus and pass strings over them, 
connected to the ring and carrying weights F^, F^ at their ends. Then by 
drawing the polygon of forces as shown we get the resultant R which should 
agree with the reading given by the spring balance. 

Experimental errors. In this and aU other experiments it 
should be remembered that it is difficult if not impossible to 
get absolute agreement between theory and experiment. This 
is due in part to the imperfections of our apparatus, introduced 



I] VECTOR QUANTITIES 13 

for example by friction in the pulleys and sagging of the 
strings due to their own weight. It is also caused by errors of 
observation ; we shall probably not transfer the directions of the 
forces to our paper without slight errors and we may make 
some mistakes in drawing our parallels to get the polygon of 
forces. The existence of these experimental errors points to 
the absurdity of trying to express the results of calculations 
based upon experimental data by numbers carried to more 
than a few significant figures. If, for instance, we express 
the weight of a girder as 7* 13762 tons we are laying claim to 
an accuracy which is impossible in practice. Manufacturers of 
steel plates, angle-bars, etc., cannot guarantee the sections 
nearer than about 2 J per cent. ; this means that if a girder is 
listed as weighing 100 lbs. per foot length, it may actually weigh 
anything between 97*5 and 1025 lbs. per foot. The above- 
cited girder therefore should be put down as having a weight 
of 7*14 tons, the remaining figures being quite meaningless on 
account of the nature of the problem. 

In this connection we would point out that it is the number 
of significant figures that matters and not the number of figures 
after the decimal point. If in the process of multipHcation we 
get a result 581,574 lbs., we should write this as 582,000 lbs.; 
or if we had -002876 foot we should call it -00288 to the third 
significant figure, that is to the same degree of accuracy as in 
the previous case. The ordinary processes of long multiplica- 
tion and division should be discarded in engineering calculations 
for logarithms or the slide -rule. 



SUMMARY OF CHAPTER I. 

The resultant of a system of forces acting upon a body is the 
single force which will have the same effect upon the body as the 
combined effect of the separate forces. 

The resultant R of two forces F^ and F2 acts tlirough the inter- 
section of their lines of action and can be found by means of measure- 
ment or calculation from a triangle two sides of which are parallel 
to and proportional to the two forces, the direction of the forces 
being maintained. This result is written R = F^ -^ F^. A force 
can be resolved in any two directions by the aid of the triangle of 
forces, but a force has no resolved part in a direction at right angles 



14 FORCES AND OTHER [ch. 

to its line of action when the directions considered are at right angles 
to each other. 

The equilibrant of a series of forces is equal and opposite to the 
resultant. 

The principle of the triangle of forces can be extended to deal 
with any number of forces, the resulting construction being known 
as the " polygon of forces.'* 



EXERCISES. I. 

We give below a number of exercises for testing the extent to 
which the reader has followed the arguments so far. The reader 
will find by experience that he learns most thoroughly by working 
the examples as he proceeds and that it is better to do a little of 
the subject thorouglily than to press forward before each step is 
mastered. 

1. Find the resultant of forces of 4 and 6 lbs. acting at 30° 
to each other. 

2. A push of 36 lbs. acts horizontally at a point upon a roof- 
truss and at the same point inclined to it at an angle of 135° in an 
anti-clockwise direction is a pull of 70 lbs. Find the resultant 
force acting upon the truss at the given point. 

3. Show that if the angle between two forces of given magnitude 
is increased, their resultant is decreased. 

4. The greatest resultant that two given forces can have when 
acting in any direction is 100 lbs. and their least resultant is 20 lbs. 
Find their resultant when they act at right angles to each other. 

5. The horizontal and vertical components of a certain force 
are equal to 5 and 12 lbs. respectively. What is the magnitude 
of the force? 

6. A nail is being driven into a vertical wall at an inclination 
of 30° to the horizontal and a man pulls the nail horizontally away 
from the wall with a force of 10 pounds. Calculate the force tending 
to extract the nail and that tending to bend it. 

7. A weight of 100 lbs. is suspended by wires from two points 
on a horizontal bar 5 feet apart, one wire being 6 feet long and the 
other 7 feet long. Find the forces in the two wires. 

8. The thrust upon the horizontal foundation of a building 
is 100 tons inclined at 10° to the vertical. Find the force or pressure 
tending to drive the foundation into the ground and that tending 
to make it slide. 



I] 



VECTOR QUANTITIES 



15 



9. An inclined force of 200 lbs. is acting upon a body resting 
upon a horizontal surface, the horizontal resistance to movement 
of the body being 40 lbs. Find the smallest angle at which the force 
can act without moving the body. 

10. A truck weighing 10 tons is resting against buffers in a 
siding the slope of which is 1 in 30; what is the pressure on the 
buffers ? 

11. Find the resultant of two forces of 15 and 36 lbs. acting 
at an angle of 52° with each other. 

12. A weight of 75 lbs. is carried by two cords which make 
angles of 35° and 51° with the horizontal. Find the pull in each 
cord. 

13. Forces 0^ = 30 lbs., OB = 60 lbs., CO = 15 lbs., DO = S0 lbs., 
OJS? = 150 lbs. meet at a point: the angles are B04 «= 45°, 
CO A = 90°, DO A = 135°, EOA = 270°. 
Find the resultant. 

14. Let Pj, P2» -Pa t)e three forces \ ,->J35 
acting at each=100 lbs. And let 
L PxOPi = 135° and L P^OP^ = 90°. 
Find their resultant. (See Fig. la.) 

15. A ball weighing 100 lbs. is 
suspended from the ceiling by a string 
8 ft. long. Find the force necessary ^^' 
to hold the weight 2 ft. from the vertical by a horizontal pull. 




f} 



CHAPTER II 

MOMENTS AND LEVERAGE 

We have considered forces up to the present only from the 
point of view of their magnitude and direction; we will now 
consider them from another point of view which is extremely 
useful in engineering problems, viz., by their moments or leverage*. 
Children accustomed to playing on a " see-saw " are aware of 
the fact that by sitting farther away from the pivot they use 
their weight to greater advantage, and that in order to get a 
balance between a heavy and a light child, the light child must 
have a greater length of plank. The scientific way of expressing 
this simple fact is that for the two forces to balance their 
moments about the pivot must be equal. 

The moment or leverage of a force about anj^ point may be 
defined as " the tendency of the force to rotate the body, upon which 
it acts, about the point. '^ It is measured by the product of the 
force into the perpendicular distance from the point to the line 
of action of the force. Referring to Fig. 11, the moment of the 
force Fi about is equal to F^ x p^. If therefore #i is 15 lbs. 
and Pi is 3 inches, the moment of Fi about = ilf ^ = 15 x 3 = 
45 pound-inches (or inch-pounds, but the former is preferable for 
a reason that we will explain later). The perpendicular distance 
from the point to tlie line of action of the force may be called the 
" arm " so that we get the general rule Moment = Force x arm. 

Positive and negative moments. Since rotation can be in 
one of two opposite directions, we must distinguish between 
these two directions by calling one positive and the other negative. 
The tendency of the force Fi is to rotate the body about in 
the direction of the hands of a clock, i.e., a clockwise direction, 
and will for convenience be called a positive moment. If the 

♦ The term " leverage " is often used in a more restricted sense than the 
above to denote tiie "mechanical advantage" (p. 53). Wo think, however, 
that it is better to use it as synonymous with " moineut." 



OH. Il] 



MOMENTS AND LEVERAGE 



17 



tendency of the force were to cause rotation in the opposite 
or anti-clockwise direction, as is the case with the force jPg, 
we should call the moment negative. It is purely a matter of 
convenience as to which is called positive and which negative. 
All that matters is that we shall agree to call positive the tendency 
to turn in one direction, and to call negative the tendency to 
rotate in the opposite direction. 

Suppose for instance that F^ = 20 lbs. and that p^, i^ 1*^ 
inches, then the moment of F2 about O = M^ = — 20 x 1*6 
= — 32 pound-inches. 




O O 



Clockwise 
Moment 



Anti-Chckwise 
Moment 



Fig. 11. Moments. 



Now so long as we are dealing with forces in one plane moments 
are scalar quantities; they have no direction in the sense that 
a force or a velocity has direction and are added by the ordinary 
or algebraic rules. 

Thus the total moment of F^ and F^ about = 45 — 32 = 13 
pound-inches. 

Moment about a point in the line of action of a force. Kemem- 
bering that a force must be considered as acting in a line rather 
than at any point, we shall see at once that a force has zero 
moment about a point in its line of action because its arm is zero 
so that the product of the force by the arm must also be zero. 

The Principle of Moments. This principle, which is of very 
great value in engineering problems and a clear understanding 
of which will obviate many difficulties that might otherwise 

A. M. 2 



18 



MOMENTS AND LEVERAGE 



[CH. 



arise, may be stated as follows : // a system oj forces in one plane 
act upon a body and keep it in equilibrium, the algebraic sicm of 
their moments about any point in the plane will be zero. 

By algebraic sum we mean the sum allowing for some being 
positive and some negative, so that we might say that the total 
positive moment must equal the total negative moment. We 
will not give a rigorous proof of this principle but will point out 
that it really follows from the idea of equilibrium and from 
Newton's first law of motion (p. 124). If a body is in equilibrium, 
it does not tend to change its state of rest so that there is no 
tendency to rotate about any point. We will restrict our 
consideration for the present to stationary bodies and will 
deal later with rotating bodies. 

Experiment 3. Verification of the principle of moments. Take a rod C 
(Fig. 12) and pivot it about one end A, allowing it to hang freely. At a 
convenient point B attach a string to the rod and pass the string over a pulley E, 




5 lbs. 



Fig. 12. Experiment on Moments. 

at a point for instance above B, and hang a weight say of 5 lbs. at the end 
of the string. Provide a convenient stop K to prevent the pull from moving 
the rod out of position. At a convenient puiut C attach another utring and 



n] 



MOMENTS AND LEVERAGE 



19 



pass it over a second pulley D and weight the end of the strmg carefully until 
the rod just begins to come away from the stop. So far as rotation about 
A is concerned the rod is then in equilibrium under the two forces 5 and W, 
there being now no pressure or " reaction " on the stop. Now measure care- 
fully the perpendicular distances from A to the two strings BE and CD; it 
should be noted that the distances are measured to the strings which represent 
the lines of action of the forces and not to the points B, C at which the strings 
are attached to the rod. 

Then taking moments about A we have 

if^ = Trx20 -5 x8=0 (because the rod is in equilibrium), 

.'. 20Tf=40 or pr=|^ = 21bs. 

W will be found to be approximately 2 lbs. in the experiment. It may not 
be exactly 2 lbs. because, as we have already indicated, there are always 
slight experimental errors especially with rough apparatus. 

The sum of the moments of a system of forces about a given 
'point is equal to the moment of the resultant about the same point. 




Fig. 13. The Principle of Moments. 



This rule can be deduced from the general principle of moments 
as follows. Let F^, F2, F^, etc. (Fig. 13) be any number of forces 
acting upon a body (in the figure we have shown four forces 
but there may be any number) and let their resultant be E. 
Now suppose that we add a force E, the equilibrant, which as 
we have shown already is equal and opposite to the resultant. 

2—2 



20 



MOMENTS AND LEVERAGE 



[CH. 



1/ 



Then the forces F^, F^, etc. and E keep the body in equilibrium, 
so that the total moment about any point must be zero: 

i.e. F^p^ + F^p^ - F^p^ + Ep^ + F^p^ = 0, 

•*• ^iPi + ■'^2^2 ~ ^aPs + -^4^4 = sum of moments of given forces 

= - ^Pb- 
But E= -R, 

••• ^iPl + F0P2 - F3P3 + ^4i'4 = - (- ^Pr) = RPR\ 

i.e. The sum of the moments of the system = Moment of resultant. 

To save writing a long string of similar quantities, it is usual 
to use the Greek letter i7 (" sigma ") to indicate " sum of quanti- 
ties Uke," so that in our present case we would write 

i:{F,p,) = Ep^. 

We shall find this notation very useful in other problems. 

Numerical verification. Take the weighted lever SQ {Fig. 14), 
which is weighted with 10 lbs. at Qj and pivoted at a point 8 inches 




^ 20 h 

Fig. 14. Forces on a weighted lever. 

from Q and is svJbjected to a horizontal pull of 20 lbs. at the point S 
the length SQ being 18 inches. The point of support or pivot of 
\& lever is called the fidcrum. If the weight of the lever itself 
nnay be neglected, the forces tending to turn the lever about the 
lulcrura are the horizontal force of 20 lbs. acting through S and 
the vertical force of 10 lbs. due to the weight acting through Q. 
j!he lines of action of these two forces intersect at the point P; 



n] 



MOMENTS AND LEVERAGE 



21 



by means of the triangle of forces we find the resultant ac of the 
forces of 10 and 20 lbs., which by measurement or calculation 
comes to 224 lbs. (i?). Draw through P a line parallel to this 
resultant. 

The perpendicular distance from the fulcrum to this line 
will be found by measurement to be 379 inches. So that 
the moment of the resultant about the fulcrum will be 
22-4 X 3*79 = - 84-8 pound-inches. 

Now take moments of the separate forces about the fulcrum. 
These are equal to - 20 x 7*07 + 10 x 568 = - 84'8 pound- 
inches as before. 



APPLICATIONS OF THE PRINCIPLE OF MOMENTS 

We will now consider some applications of the principle of 
moments to practical problems. Further applications will arise 
at later portions of the book. 

Reactions on a beam. Determine the reactions on a beam 
of 20 feet span loaded in the manner shown in Fig. 15. By " re- 
actions " on a beam we mean the pressure exerted by the support 



A A 






U' 



w, 



y. 

lb 



iWc 



1 2 



■^- 



3 — >H- 






4' 



i^ 



B 



Fig. 15. Reactions on a beam. 



upon the beam. If the support is statioimry, it will press upward 
upon the beam with a force equal to that with which the beam 
presses against the support. This ifi an example of Newton's 
Third Law of Motion that " Reaction la equal to action." This 
*' reaction " may be regarded as a force induced to counteract 
the original force (or " action ") so as to bring the resultant 
force at the point to zero. 

In our present example there are six forces acting on the 
beam which keep it in equilibrium, viz., the four weights Wi,W2, 
Wq and 1^4 and the two reactions i2^ and E^. 

In the present case our forces are all vertical, the weights 
acting downwards and the reactions upwards. We have seen 
that, for equilibrium, the vector sum must be zero. When 



22 



MOMENTS AND LEVERAGE 



[CH. 



the forces are all in the same direction, the vector sum is the 
same as the algebraic sum so that we get the rule that " the 
total upward force must be equal to the total downward force." 

Therefore we have : 
i2^+ ^^5= T^i + T^2 +F3 + 1^4 = J + i + 1 + 2 = 3£- tons. 

To determine one of the reactions, say Ei, take moments 
about the other support B. The moment of R^ about B is 
zero because the " arm " is zero. 

The algebraic sum of all the moments about B must b« 
zero, i.e. 

iJ^ X 20 - Tfi X 17 - TF2 X 13 - Tf 3 X 7 - TF4 X 4 = 0, 



I.e. 



20Rj_ =17xi+13xJ+lx7 + 2x4 
= 26'75 tons-feet. 



R.= - 



26-75 



20 



1-34 tons nearly. 



.-. Rs= 3-75 - 1-34 = 2-41 tons. 

As a check on the working we will now find Rg by taking 
moments about A. We then have 
20Rs= ix3 + ix7+lxl3-}-2xl6 
= 48*25 tons-feet. 

48-25 
20 



R. 



= 2-41 tons nearly. 



X 



A 



B 



Stability of a wall. A wall 18 ins. 
thick and 8 feet high weighs 6 tons. 
Find what horizontal 'pressure, due to 
the wind acting ai the centre of the wallf 
would be necessary to overturn the wall. 
Referring to Fig. 16, the forces acting 
upon the wall are the horizontal wind 
pressure P and the weight W which 
may be taken as acting down the centre 
line of the wall. 

Taking moments about the point B 
we have a clockwise moment of P x d 
called the " overturning moment " and an anti -clockwise moment 
of W X X called the " stability moment." If the overturning 
moment is less than the stability moment, the wall will not 
overturn, but if the overturning moment is ever so little greate** 



W 
Fig. 16. Stp.bility of a ^all. 



n] 



MOMENTS AND LEVERAGE 



23 



than the stability moment the wall will topple over. In the 
limiting case in which the two moments are equal, the wall is 
just about to overturn, and the value of P when the moments 
are equal is usually taken as the least value required to cause 
overturning. In this case therefore we have 

Pxd:=Wxx, 

or p = — - . 

a 

Now TF = 10 tons. (Z = 4 ft. and a; = -^ ft. 

.*• P = -A rs = 1*875 tons. 

4 X 12 

The Lever Safety Valve. The lever safety valve provides 
another common example of a device in which the principle 




I I I 1 I I I I I 



¥ 



w 



^ 



Fig. 17. Lever Safety Valve. 

of moments is used to facilitate calculations. The device consists 
of a lever AO (Fig. 17), pivoted* at 0, and provided with a weight 
W the position G of which is capable of adjustment along the lever. 
A force P due to the pressure of steam in the boiler acts at the 
point B and the parts are so proportioned that when the pressure 
of steam in the boiler reaches a pre-determined limiting value of 
safety, the force P is sufficient to hft the lever and allow the 
steam to escape until the pressure falls to the required extent. 
There are four forces acting upon the lever: the upward 

* The point about which a lever is pivoted is commonly referred to as thn 
■"Mcrum." 



24 MOMENTS AND LEVERAGE [ch. 

force P acting through the point B; the movable weight W 
acting downwards through the point C; the weight w of the 
lever itself acting through a point G called the " centre of gravity," 
the position of which is found in the manner which we shall 
describe later; and finally the downward reaction R caused 
by the lever pressing upwards upon its pivot or fulcrum. In 
calculations we want to know the pressure P required to lift 
the valve for a given position of the weight W upon the lever. 

By taking moments about the fulcrum we get rid of the 
force R because its moment about a point in its line of action 
is zero. We will show later that we might have taken moments 
about any other point but that it would take longer. 

Then we have : Clockwise moment about O = Wx + wb ; 
Anti-clockwise moment about = Pa; 

:.Pa=Wx + wb (1), 

p^Wx±wb 

a ^ ' 

Numerical Example. Take, TF = 62 Ihs.y w = VI lbs., 
X = S3 ins., b = 16i ins., a = 3J ins. If the diameter of the valve 
be 3 ins. find the pressure in the boiler at which the safety valve 
will " blmo off.'' 

Let p lbs. per sq. in. be the pressure required. 

Area of valve = ^ x 3^ = 7 07 sq. ins. 

.*. Total pressure P = p x area = 7-07p. 
Putting the values in (2) we have 

62 X 33 + 12 X 16-5 2244 



7-07p = 



3-25 3-25 



sr » 



2244 
•■• P = 707 X 325 = 97-7 lbs, per sq. in. 

Calibration of Safety Valve. By *' caUbration '* of an 
instrument is meant the determination of the scale for measuring 
the quantities with which the instrument deals. Suppose that 
we wish to mark points along the lever of the safet}'^ valve which 
we considered in the previous example to correspond to pressures 
from 60 to 100 lbs. per sq. in., rising by 5 lbs. per sq. in. at a time. 
Suppose that when the pressure is p lbs. per sq. in. the lever is in 



11] 



MOMENTS AND LEVERAGE 



25 



equilibrium with the weight w at a distance x from the point O. 
We have seen that P = lOlp. 

Therefore taking moments as in equation (1) we have 

707^9 X 3-25 = 62a; + 12 x 16'5, 
.-. 23^ = 62a; + 198, 
/. 23i) - 198 = 62a;, 
/. X = {Zn^p - 3-19) inches (3). 

Equation (3) is what is called a " linear " equation because if 
values of the lever-length x be plotted against the pressure 'p 
the resulting diagram or graph will be a straight line. 

40 



(0 



30 



S 20 



^_ 



^"^QO 70 60 90 IQO 

PieBSure in lbs. per sq. in. 

Fig. 18. 

Fig. 18 shows the graph which is the " calibration curve " 
of the safety valve. 

For ^ = 60 lbs. per sq. in., x = -371 x 60 - 3*19 = 19-1 inches. 
For p = 100, we get x = 37*1 - 3*19 = 33'9 inches. 

The distance therefore between the 100 and the 60 marks on 

the lever is 33'9 — 19*1 = 14-8 inches and we have seen from 

the diagram that the divisions are equally spaced. Therefore 

the length of each of the eight divisions corresponding to 5 lbs. 

14-8 
per sq. in. will be — ^- = 1*85 inches. 

o 

Equilibrium of a body under three forces. There are a 
number of problems in which the number of forces acting can be 



26 



MOMENTS ANT> LEVERAGE 



[CH. 



J reduced to three. We can then make use of the following rule. 
1/ // three forces act upon a body and keep it in equilibrium y they 
must be in one plane* and are either parallel or their lines of action 
meet at a common point. We will first prove the rule by assum- 
ing that it is not true and that three forces F^, F^, F^ (Fig. 19) 
act upon a body and keep it in equilibrium. Let the lines of 
action of F^ and F^ meet at A and suppose that the line of action 
of ^3 does not pass through A. Now take moments about A, 
The moments of F^ and F2 about A are each zero, but F^ has a 
moment of i'^g x ^ so that the sum of the moments of the three 
forces about A is equal to F^x p\ we have seen, however, that 



<j 




Fig. 19. 

if a body is in equilibrium the sum of the moments about any 
'point of all the forces acting upon it is zero. The only possible 
way for the total moment to be zero in the present case is for 
i^3 also to pass through A. If two of the forces are parallel, the 
third must also be parallel to them; if not the third force will 
intersect one of the others at some point and there would be a 
resultant moment about that point due to the third force. The 
student should draw his own diagram to illustrate this. 

This fact of the concurrency of three forces which keep a 
body in equilibrium is employed in several problems involving the 
reactions at the supports of structures. As a simple illustration 
take the case of a lever AB (Fig. 20) pivoted at the lower point B 
and held in the inclined position shown by a horizontal force P 
acting through the point A. The three forces acting upon the 
lever are the weight W acting vertically through the point G, 

♦ The general consideration of forces not in one plane lies outtdde the scope 
"•f this book. One simple case ia dealt with on p. 184. 



"] 



MOMENTS AND LEVERAGE 



27 



the horizontal pull P and the reaction R at the pivot at B. The 
weight and pull act in given lines which intersect at the point O ; 
the reaction R must therefore also pass through 0, so that by 
joining OB we get the line of action of the reaction R. We have 
now the directions of three forces in equilibrium, but have still 
to determine their magnitudes. This is effected by drawing 
the triangle of forces as already explained, setting down 1, 2 to 




represent the weight W to convenient scale ; through 2 draw a line 
parallel to the force P and through 1 draw a force parallel to the 
reaction R. The intersection 3 gives the third point required 
in the triangle, and 2, 3 will represent the force P and 3, 1 the 
force R to the same scale as that to which 1, 2 represents W. 

Graphical construction for moments; link and vector 
polygon construction. We have dealt already (p. 10) with the 
graphical construction for finding the magnitude and direction 
of the resultant of a number of forces. We now come to a»ii 
extension of that construction. 



28 



MOMENTS AND LEVERAGE 



[CH. 



Let 0, 1 ; 1,2 (Fig. 21), and so on, be a number of forces 
not necessarily parallel nor concurrent. To some suitable scale 
set down on a vector figure 0, 1, 2, and so on, then as before, 
the closing line 0, 6 gives the magnitude and direction of the 
resultant. Now take any point or pole P at any convenient 
position on the paper and join P, 0; P, 1 ; and so on. Then 
draw anywhere across the line of action of the first force a line 
af parallel to P, and cutting the line of action of the force 
in a ; across space 1 draw ab parallel to P, 1 ; across space 2, 
draw cd parallel to P, 3, and so on until the last line or Unk parallel 
to P, 6 is reached. Produce this last link to meet the first 
link in /, then the resultant E will pass through the point /, 



9 



\'^,S Link Toly son I 





l/ecTor folys^" 
Fig. 21. Link and Vector Polygon construction. 

and the figure a, b, c, d, e, f is called the link polygon or by some 
writers the funicular polygon. 

Suppose the moment of the given force system is required 
about the point Q. Through Q draw a line parallel to the 
resultant B to cut the first and last links produced in h and g. 
Then if the point P is at perpendicular or polar distance p from 
0, 5 on the vector figure, the moment of force system about Q 
is equal to gh x p, gh being read on the space scale and p on 
the force scale. 

Proof. By the law of vector addition, the force 0, 1 on the 
vector figure is equivalent to forces OP. PI acting in fa and ab; 
the force 1, 2 is equivalent to forces IP, P2 acting in ba and be, 
and so on, the last force 4, 5 being equivalent to forces 4P, 
P5 acting in de and fe. It will be seen that with the exception 
of the forces down fa and fe all these forces neutralise each other. 



n] 



MOMENTS AND LEVERAGE 



29 



and so the resultant of the whole system of forces is the same 

as that of ja and /e and therefore acts at the point of intersection 

/ of these forces. 

The triangles jgh, P, 0, 5, are similar, the corresponding sides 

being parallel. 

. ^ ^ 0, J 

(because in similar triangles the bases are proportional to the 
heights), 

:. p X gh = 0, 6 X q, 

but 0, 5 = resultant R and q is distance of R from Q^ 

.*. 0, 5 X gf = moment of force system about Q, 
:. p X gh = moment of force system about Q. 

Numerical Example. Find the resultant of the loads shown 
in Fig. 22 and find their moment about the point A. This is the 



Polygon 




Vector 
Figure 

Kg. 22. 

same system of loading as we considered analjrtically in the 
example on p. 21. 

Number the spaces 1, 2, 3, etc. between the forces, and choosing 
a convenient scale of loads set down the vector figure 1, 2, 3 . . 5, 
which is a vertical straight line in this case because ail the 
forces are vertical. Then, choosing any convenient pole P, 
join P to 1, 2 . . 5, draw anywhere across the first force line 



30 MOMENTS AND LEVERAGE [ch. 

a line af parallel to PI, cutting it in a; then across space 2 
draw ah parallel to P2 ; across space 3 draw be parallel to P3 ; 
across space 4 draw cd parallel to P4; and through d draw 
dj parallel to P6 cutting the first link af in /. Then the resultant 
R, which is equal to 1, 5 on the vector figure, i.e., 3*75 tons, 
acts through/ which is at a distance 12*9 ft. from the left-hand end 
A of the beam. To find the moment of the system of forces 
about A draw a vertical through A and let fa produced meet 
it in g and df produced meet it in h. Then, according to the 
construction proved above, gli x 'p, i.e., 19*3 x 25 = 4825 tons- 
ft., is the moment of the system of forces about A. 

Extension of construction to find reactions. If we want to 
find the upward reaction of Rb at the support B we could then 
divide as before this moment by the " arm " of the reaction, 
i.e., by 20. We can, however, extend the construction as follows 
to do this division graphically. Let the last link fdy produced, 
cut the vertical through the support j5 at a point e. Join gre, 
as shown in dotted lines, and through the pole P draw a line 
Px parallel to ge to cut 1 , 5 in a;, then 5x will equal the reaction 
Rb and x\ will equal the reaction jB^ (because as we have seen 
already the sum of the reaction must be equal to the total load 
if the beam is in equilibrium). 

Proof. The triangles Pxb and egh are similar because their 
corresponding sides are parallel. 

/. since their bases are proportional to their heights we have 

x^ _gh ^ 

• a:5-^- ^- 

moment of force system about A 
= I . 

But we have by the principle of moments that 

Rji X I = moment of force system about A, 

n moment of force system about A 
I.e., Rb = j^-^ ; 



x,5 = R 



B' 



Alternative proof. We can prove this in a similar manner 
to that on p. 28 as follows: W^^ 1, 2 can be replaced by its 



n] 



MOMENTS AND LEVERAGE 



31 



components IP in ga and 2P in ba\ Pf g = 2, 3 by its components 
2P in ab and P3 in cb and so on. Then P2 in ba balances 2P 
in ab and so on, so that we are left with only IP in ga and P5 
in ed. These meet in / so that the resultant R passes through /. 

Also IP in ga can be replaced by components xP in ge and 
\x in Ag, and P5 can be replaced by Px in eg and a:5 in Be. The 
forces in ge balance and the system reduces to \x down at A 
and a;5 down at B and these forces are equal and opposite to 
the reactions at A and B. 

We will deal further with this construction later in considering 
calculations for beams and girders. 

Couples. When the forces acting upon a body reduce to 
two equal and opposite parallel forces, they are said to form 
a couple. Thus the forces F (Fig. 23) form a couple. A couple 




Fig. 23. Couples. 

has no resultant because the vector sum of the forces composing 
it is zero (i.e., F — F = 0); but about any point O in the plane 
of the forces the couple has a moment equal to F . a (a is the 
perpendicular distance between the forces and is called the arm). 



32 



MOMENTS AND LEVERAGE 



[CH. 



To prove this take moments about the point O. Then we have 

Mo = F X BO -F xCO = F{BO - CO) = F . EC = F .a. 

The effect therefore of this couple v/ill be to rotate the body 
upon which it acts, without moving the body as a whole. The 
couple shown is clockwise. 

The only way in which a couple can be neutralised or equili- 
brated is by the introduction of another couple of equal moment 
but opposite direction. The vector sum of the forces will still be 
zero and the moment about any point will also be zero, both 
conditions of equilibrium being therefore satisfied. 

We have an example of couples in the tool shown in Fig. 24 
diagrammatically, for enlarging holes in wood. The tool has 



R 




R 




Fig. 24. Couple acting upon cutting tool. 

two cutting lips and is operated by a lever AB which is grasped 
by the operator's hands. At C and Z>, the points of contact 
of the cutting lips with the wood, resisting forces R are brought 
into play tending to prevent the tool from rotating. 

The operator exerts a couple of moment P x I and the 
resistance exerts a couple of moment R x a. When the operator 
starts to press on the lever, it does not move ; it is then in equi- 
librium and P X I the moment of the operative couple is equal 
to R X a the moment of the resisting couple. The operator 
then presses more strongly until the tool moves round, i.e., 
until he has exerted a couple greater than the resistance of the 
wood can exert. 

Conditions of equilibrium in link and vector polygon con- 
struction. We have seen for a system of forces to be in equilibrium 
the conrlitions to be satisfied are (a) that the resultant is zero, 
(6) that the total moment of all the forces about any point must 
be zero. In the js^aphical construction, {a) is satisfied if the 
first and last points of the vector polygon coincide, i.e., if the 



n] MOMENTS AND LEVERAGE 33 

vector polygon closes. To satisfy condition {b) the distance gh 
in Fig. 21 must be zero for every position of the point Q, in other 
words g and h must coincide. The only way in which this can 
happen is for the first and last Hnks to coincide, or for the link 
polygon to close. If the vector polygon closes but the link 
polygon does not close, the system reduces to a couple. Since, 
as we have seen, a couple has the same moment about any point, 
we should expect that the first and last links must be parallel. 
If we consider the construction we see that this must be the 
case because the first and last points 0, 5 of the vector polygon 
coincide. Therefore the first and last links will both be drawn 
parallel to P, and must therefore be parallel to each other. 



SUMMARY OF CHAPTER II. 

The moment of a force about any point is measured by the 
product of the force by the perpendicular distance from the point 
to the line of action of the force. 

If a system of forces in one plane act upon a body and keep it 
in equilibrium, the algebraic sum of their moments about any point 
in the plane will be zero. 

The sum of the moments of a system of forces about a given 
point is equal to the m.oment of the resultant about the same point. 

If three forces act upon a body and keep it in equilibrium, they 
are in one plane and are either parallel or their lines of action meet 
at a common point. 

When two eqvial forces are parallel and opposite in direction 
they are said to form a couple. The moment of the couple is measured 
by the product of one of the forces by the perpendicular distance 
between them. And a couple can be neutrahsed only by the intro- 
duction of another couple of equal moment but opposite direction. 

The link and vector polygon construction enables us to determine 
graphically the position, direction and magnitude of the resultant 
of a number of forces and the moment of the resultant about any 
point. 

If the system is in equilibrium, the Jink and vector polygons 
are both closed ; if the system reduces to a couple the vector polygon 
is closed but the first and last links of the link polygon are parallel. 

A. M. 3 



34 



MOMENTS AND LEVERAGE 



[CH. 



EXERCISES. IT. 

1. Define the moment of a force. A lever AB 13 hinged at A 
and carries weights as shown (Fig. II a). What force P acting 
upwards will keep the bar in a horizontal position? 

2. A lever AB whose weight is 120 lbs. and length 3 feet has 
a fulcrum 10 inches from the end A. What weight at the end B 
will balance 384 lbs. placed at A ? Also find the pressure on the 
fulcrum. 

3. A uniform lever 26 inches long and weighing 45 lbs. carries 
a weight of 20 lbs. at one end and 35 lbs. at the other. Find the 
point in the lever about which it will balance. 

4. A uniform lever 8 ft. long weighs 42 lbs. It carries a weight 
of 36 lbs. at one end and 24 lbs. at the other. Find the point about 
which it will balance. 



K- 10 



2^ 



-16 



12 lbs. 
Fig. II a. 



5lb3. 




Fig. 116. 



5. A safety valve is 3 inches in diameter and the weight on the 
end of the lever is 55 lbs., the distance of the fulcnmi from the centre 
of the valve being 4'5 inches. If the weight of the lever and valve 
are negligible, how far along the lever from the centre of the valve 
must the weight be placed if the valve is to blow oS at a pressure 
of 80 lbs. per sq. inch ? 

6. The area of a safety valve is 8 square inches. The lever is 
2 ft. 6 ins. long, its centre of gravity is 1 ft. from the fulcrum and 
its weight is 10 lbs. The fulcriun is 4 inches from centre line of 
valve. Find the pressure in lbs. wt. per sq. inch at which steam 
will blow off, if the weight on end of lever is 66 lbs. and the valve 
itself weighs 1^ lbs. 

7. A lever 16 inches long weighs 25 lbs. and has a fulcrum at 
one end. It is held in a horizontal position by a vertical force 
applied at the other end. The lever being uniform, what is the 
magnitude of this force? 

8. A uniform lever AB whose weight of 15 lbs. acts at the centre 
is 15 inches long; it is hinged at A and held horizontally by a cord 



n] 



MOMENTS AND LEVERAGE 



35 



carrying a weight W as shown (see Fig. II 6). Find the magnitude 
oiW. 

9. A pole AC pivoted at C and carrying a weight of 1 ton is 
supported by a rope AB. Prove that the pull in AB will be least 
when its direction is at right angles to AG» Find this pull, and 
the thrust in the pole. (See Fig. lie.) 

10. A system of five parallel forces whose magnitudes are 10, 
12, 8, 6, 11 lbs. weight respectively act in lines 2 ins. apart. Find 
the position of their resultant. 

11. A bent lever ACB is pivoted at G ; the arm AG is horizontal 
and 9 inches long; the arm BG is vertical and 39 inches long. A 
load of 300 lbs. is hung from A. Find what horizontal force at B 
will produce equilibrium, neglecting the weight of the lever. 

12. A beam 20 ft. long supported at its ends has a load of 2 tons 
at the centre of the span, another of 1 ton at 3 ft. from one end, 
and another of 3 tons at 4 ft. from the other end. Find the reactions 
of the supports neglecting the weight of the beam. 




!<. 6ft 

I 
I 



---^ 



-9B 



6 
50 lbs. 



!<. 5ft ^ 




Fig. lie. Fig. lid. 

13. Fig. lid shows a compound lever. The fulcra are at C 
and E. Find the weight W which can be supported by an effort 
of 50 lbs. applied as shown, neglecting the weight of the beams. 

14. Forces of 1, 2 and 3 lbs. are parallel and act at the corners 
of an equilateral triangle. Find where the resultant acts. 

15. In an 8-oar boat each man pulls with a force of 60 lbs. 
If the oars are 10 ft. long and 2 ft. 6 ins. from hand to rowlock, 
find the force impelling the boat forward. 

16. If a balance has unequal arms a and h and a shopman 
weighs alternately from each scale pan, does he ultimately lose 
or gain and how much? 

17. If the span of a beam is 20 ft. and a load of 12 cwt. is shifted 
from one position through a distance of 5 feet along the beam, what 
difference in the reactions will this cause? 

3—2 



CHAPTER III 

WORK, POWER AND ENERGY 

The term work is quite familiar to everybody but its general 
meaning is not very easy to express succinctly; it is used in 
mechanics in a special restricted sense and may for our purpose 
be defined as follows : When a force acts upon a body and catises 
it to move it is said to do work on the body. When the force is 
constant, work is measured by the product of the force and the 
distance through which the body moves in the direction of the 
force. The engineer's unit of work is the foot-pound, i.e., 
the amount of work done by a constant force of one pound 
weight in moving a body through a distance of one foot in the 
direction of the force. If our forces are measured in tons and 
our distances in inches, our work will be in inch- tons and so on. 
If for instance the weight used in driving a clock weighs 20 lbs. 
and it drops through a vertical distance of 5 feet, the distance 
moved in the direction of the weight of the body, which acts 
vertically, is 5 feet. Therefore the work done by the weight 
is 20 X 5 = 100 ft.-lbs. 

To express this idea generally, instead of by numerical 
illustration, suppose that a constant force F (Fig. 25) acts upon 
a body indicated by the shaded area situated originally at a point 
A and that after a certain time the body has been moved to a 
point B. Draw BG parallel to F and draw ^C at right angles 
to it, then BC is the distance through which the body has moved 
in the direction of the force, so that the work done in the given 
time is measured by i^ x BC. 

It will be noticed that the work done by a force is measured 
by the product of a force into a length and that the moment 
of a force about a point is also measured by the product of a 
force into a length. In order to avoid confusion a distinction is 



OH. ni] 



WORK, POWER AND ENERGY 



37 



sometimes made in naming the compound units — thus work is 
measured in foot-pounds, inch-tons, etc., and moments are 
measured in pound-feet, ton-inches, and so on. 

It is very important at this stage to note that if the force 
is given in magnitude and direction the work done in the given 
time depends only upon the original and jfinal positions, A and 
B respectively, of the body ; it does not depend at all upon the 
path taken. The work done, for instance, for a straight line 




Fig. 25. 

path between A and B is the same as for any curved path 
such as that shown in the figure. 

We will here note that the product of the force and the 
distance moved in the direction of the force is exactly equal 
to the product of the resolved part of the force in the direction 
of motion and the actual straight distance moved. To prove 
this statement draw CD perpendicular to AB^ and suppose 
that a scale of force is so chosen that CB represents the force F. 
Then DB represents the resolved part of F in the direction of 
movement AB. 

Then the work done = F x BO = BG^, 



3B WORK, POWER AND ENERGY [ch. 

BC 
Now in the A ABC, sin 6 = -^-5 , and in the A DCB, 

but DB is the resolved part of F in the direction of motion and 
^5 is the distance moved so that the work done may also be 
measured by the product of the resolved part of the force in the 
direction of motion and the straight distance moved. Suppose 
for instance that F is 10 lbs. and AB is 6 inches and the angle 6 
is 30°. Then BC will be 3 inches and the component of i^ in 
the direction AB will be 5 lbs. So F x BC = ^0 inch-lbs. 
and 5 x AB = 30 inch-lbs. 

Movement of the body in the direction of the force is essential 
for work in the scientific sense. If a man is standing still and 
is holding a heavy body, he must be exerting by muscular action 
a force on the body equal to the weight If lbs. of the body if 
it is also stationary ; but he is not doing any work on the body 
in the scientific sense, although he would probably feel aggrieved 
if told so. If, however, he lifts the weight through a certain 
vertical distance x feet, his muscular effort does work to an amount 
Wx ft.-lbs. He does the same amount of work whether he lifts 
the weight straight up or in an inclined or curved path; he 
also does the same amount of work whether he lifts the weight 
quickl}'- or slowly. 

Power. If he lifts quickly, however, he exerts more power 
than if he lifts slowly, for power is the rate of doing work, i.e., 
power is the work done in a unit of time. The British unit 
of power is called the Horse-Power (h.p.) and is fixed at 33,000 
ft.-lbs. per minute (or 550 ft.-lbs. per second). This unit was 
chosen by James Watt as the result of experiment with horses 
winding up weights. It is not a very satisfactory unit but it 
has become firmly estabhshed and it is now too late to alter 
it. [We have already used the idea of a 1000 lb. unit of force 
called the kip, so that a 1000 ft.-lbs. would be called a foot- 
kip. If we were setting out to choose a more convenient unit 
of power than the horse-power, we might take one foot-kip 
per second and call it the Skip which would be equivalent to 



Ill] WORK, POWER AND ENERGY 39 

-Wit = 1*8181 h.p. We do not propose, however, to adopt this 
unit throughout the book.] 

Numerical Examples. {I) What is the least po2ver that a pump ^ 
must he exerting when it is lifting water at the rate of 1500 gallons 
per minute through a vertical distance of 50 feet ? 

A gallon of water weighs 10 lbs., so that the force exerted 
on the water is at least 10 x 1500 = 15,000 lbs. Actually the 
force will have to be a little more than this because of frictional 
and other resistances that have to be overcome; that is why 
the question is worded in the form given. 

The work done per minute = 15,000 x 50 

= 750,000 ft.-lbs. 
750,000 
••• ^^^"" = -33:000 
= 22-7 H.p. 



(2) // the horse referred to in the example on p. 7, travels at 
the rate of 3 miles per hour for 10 minutes, how much work will 
he have done and at what horse-power will he he working? 

In ten minutes the horse will have walked half a mile, that 
is 2640 feet. 

We have shown already that the component of the pull 
in the direction of motion of the horse is 141 lbs , so that the 
work done is 141 x 2640 = 372,000 ft.-lk^. One f.p. = 33,000 
ft.-lbs. per min.= 330,000 ft.-lbs. in 10 minutei. 

•■• Ho^««-Po^«r = 11?^ ™ 103, 

Energy. Energy is usually defined a«i tlit- capacity for 
doing work. Energy exists in nature in several forms; thus 
wt) have electrical energy, light energy, heat energy, energy 
stored in water at high elevations, the principal source of all 
energy being the heat of the sun. Energy can be converted 
from one form into another and the principal function of the 
engineer is the conversion of natural eri'^rgy into convenient 
forms for the benefit of man. Thus tho energy stored up by 
the sun in bygone ages in the vegetation T<hich has now become 
coal is converted in the boiler into the expansive energy of 
steam which drives steam-engines for performing countless kinds 



40 WORK, POWER AND ENERGY [ch. 

of mechanical work. In mechanics we divide mechanical energy 
into two kinds — kinetic energy and potential energy. 

Kinetic energy (commonly written k.e.) is the work which a 
body is capable of performing in virtue of its motion. A familiar 
exaiiiple of kinetic energy is that possessed by a bullet, which 
in being brought to rest can do a large amount of work ; another 
example is that possessed by the wind, the kinetic energy of 
which has been employed from time immemorial to propel ships, 
and drive mills for the grinding of com and for the pumping 
of water. 

Potential energy is the work which a body is capable of performing 
in virtue of its position. A familiar example of this is given in 
an illustration which we have already considered, viz., a weight 
used in driving a clock. If the weight TFlbs. is at a height 
h feet above the ground, it will do an amount of work equal to 
Wh ft. -lbs. before it comes to the ground ; its potential energy 
is therefore said to be equal to Wh ft. -lbs. Another example 
is afforded by water at a high elevation which is often used 
to drive machinery to generate electric power. 

In this connection we may point out that when in ordinary 
parlance we speak of power we really mean energy. An electric 
power station is really employed in generating electric energy, 
power meaning strictly, as we have already indicated, the rate 
of doing work. 

The Conservation of Energy. We have already stated 
that energy can be converted from one form to another, but up to 
the present nobody has discovered a way of creating or destroying 
energy, nor do we think it probable that anybody ever will. 
Thus we get the doctrine which is the foundation of all physical 
science that " Energy can neither be created nor destroyed 
but can be converted from one form into another." This is 
known as the principle of the " conservation of energy." Failure 
to understand this law has led thousands of men to spend 
much valuable time and money in trying to invent perpetual 
motion machines, i.e., machines which when once started will 
go on working for ever without receiving any additional external 
energy. It is really quite remarkable that, although this doctrine 
has long been accepted by all scientists, there are still inventors 
who try to cheat nature of her laws and to make these machines. 



Ill] WORK, POWER AND ENERGY 41 

As no energy is destroyed a given amount of heat energy can 
always be converted into the same amount of work. Thus one 
British Thermal Unit (b.th.u.), which is the amount of heat 
required to raise the temperature of one pound of water through 
one degree Fahrenheit, has been found to be equivalent to 
778 foot-pounds of work. This is commonly spoken of as the 
Mechanical Equivalent of Heat or Joule's Equivalent. 

The commercial imit of electrical energy in this country 
is called the Board of Trade Unit (b.t.xj.) and the unit of electrical 
power is the watt or the kilowatt (1000 watts) ; 1 kilowatt is 
one B.T.u. per hour (i.e., 1 kilowatt is the power which in one 
hour produces one b.t.u. of energy). Now 1 Horse-Power is 
equivalent to 746 watts; when therefore we know the amount 
of electrical energy used in a given time in any machine such 
as an electro-motor, we can tell exactly how many foot-pounds 
this is equivalent to. 

Useful Energy. While it is true that energy cannot be 
destroyed, it is also true that in every conversion from one form 
of energy to another some of it is always wasted. Take for 
instance the case of a steam, gas or oil engine. A certain amount 
of energy is put into the engine in the form of steam or explosive 
mixture and a certain amount of work (called useful work) is 
done by the engine, but we can never use more than something 
like one-quarter of the amount of energy put in ; of the remainder 
part escapes in the exhaust steam or gases and part is spent 
in overcoming the friction in the engine. The energy is not 
destroyed but much of it cannot be usefully employed; it is 
practically wasted. One pound of average coal contains about 
12 million foot-pounds of energy ; the greater proportion of this 
goes into the water in the boiler and the remainder goes up the 
chimney. It is a very good steam-engine that does not use 
more than 1| lbs. of coal per h.p. hour. 

Now 1 H.p. hour = 33,000 x 60 ft.-lbs. 

= 1-98 million ft.-lbs. 

\h lbs. of coal contain IS million ft.-lbs. of energy so that in 
a very good steam-engine 1*98 out of 18 or 11 per cent, of the 
energy supplied to it is usefully employed ; the remaining 89 per 
cent, is wasted. The great problem to be faced by engineers 



42 



WORK, POWER AND ENERGY 



[CH. 



of the future is that of obtaining mechanical energy in a less 
wasteful manner. An electro-motor wastes very much less 
energy than a steam-engine, but the electrical energy is nearly 
always obtained from coal by means of a steam-engine so that 
the energy of the coal is still to a large extent wasted. Gas 
and oil engines have now become less wasteful than steam- 
engines but even the best of them cannot give out as useful 
work more than one-third of the energy supplied to them. 

We shall return to this subject of wasted energy in the 
next chapter. In the meanwhile we will emphasize the fact 
that the conservation of energy is to be the basis of our treat- 
ment of mechanics. In any operation of a machine or action 
of a number of forces we will endeavour to find out what has 
become of the work that has been performed and by drawing up 
a kind of work balance-sheet we shall be able to investigate a 
number of points which are of the utmost importance in practice. 

Work done by a variable force. In our examples illus- 
trating the idea of work we have considered up to the present 
only the case in which the force is constant, but in most cases in 



o 

O 




H MEG 

Distance in direction of force 

Fig. 26. Work done by a variable force. 



practice the force varies from one time to another and if we 

based our ideas upon constant forces only we should not be 

able to deal clearly with the problems that arise in practice. 

Suppose that the force acting upon a body in the direction 



Ill] WORK, POWER AND ENERGY 43 

of its motion and driving it forward varies so that when we 
plot a diagram of the force at various points we get a curve 
ABCD, Fig. 26, which is usually called the effort curve. Consider 
two points E and G which are so close together that the force 
F may be regarded as constant over the length. Then the work 
done over this length will be equal to constant force x distance 
moved = F x S = area of the shaded strip of the curve. The 
reader will see that the smaller we make the distance EO the 
more nearly true will be the statement that the area of the strip 
is equal to F x S, but that for comparatively long lengths the 
statement is only approximate. If now we consider the whole 
base HJ to be divided up into short lengths, the same argument 
will hold for each strip of the curve so that adding together 
these separate strips we see that the total work done in moving 
the body from H to J is represented by the area HAD J. 

Or we get the rule that : 

The work done is represented by the area beneath the effort 
curve. 

Now suppose that we wish to find the work done up to various 
points along the base and to obtain a diagram representing to 
some other scale the work done. Such a diagram will be of the 
form shown by the curve HLP in the figure and is called the 
worh curve. Consider any point L on this curve. Then the 
ordinate LM represents the work done in moving from H to M 
and this is also given, as we have proved above, by the area 
HAKM. We see therefore that the ordinate of the work curve 
at any point represents the area of the effort curve up to the 
same point. When two curves have this relation, the first is 
said to be the sum curve of the second; thus in our case the 
worh curve is the sum curve of the effort curve. A graphical con- 
struction for the sum curve is given in the appendix (p. 294). 

Numerical Example. The force urging a body forward 
increases uniformly from zero to 2000 Z65. during the first 15 feet 
of movement; it then remains constant for the next 20 feet; and 
finally decreases uniformly to zero in a further 20 feet. Find 
the work done and the constant force which would do the same 
amount of loork in moving the body through the same distance. 

Fig. 27 shows the effort curve in this case. Therefore the 
work done is given by the area of the figure ABCD. 



[CH. 



44 WORK, POWER AND ENERGY 

This is equal to 

ft.-lbs. 

Area of A ABE = J x 15 x 2000 = 15,000 

Area of rectangle BCFE = 20 x 2000 = 40,000 

Area of ACDF = J x 20 x 2000 = 20,000 

Total work done = 75,000 ft.-lbs. 




-> Distance 



Kg. 27. 



The total distance moved is 55 feet. If therefore a constant 
force F were acting we should have 55F = 75,000 ; 

„ 75,000 .oaA M. 1 

.*. F = — KR— = 1364 lbs, nearly. 

We show in dotted lines in Fig. 27 the work curve for this 
case. The portion AH is a parabola with vertex at A ; HJ is a 
straight line, and JK is a parabola with vertex at K. The 
student should make this construction as an exercise. Take for 
instance as scales : Distance V = 10 feet. Force V = 1000 lbs. 
Polar distance p = 2-5 actual inches. 

Then the work scale will be 1'' = 2-5 x 10 x 1000 = 25,000 
ft.-lbs. 

DK should therefore be 3 inches. 

Work against Resistance. In every case that arises in 
practice there is a force resisting the movement of a body imder 
the driving force or effort, such resisting force is called the 
Resistance or sometimes the "external resistance." 



in] 



WORK, POWER AND ENERGY 



45 



Take the case of a steamboat, Fig. 28. The steam acting 
upon the pistons and thence upon the propeller causes a certain 
tractive effort F to be exerted tending to push the steamer 
forward; the resistance of the water and other external forces 
tending to resist the forward motion of the steamer cause 
a resistance force S to be exerted in the opposite direction. 
If F is greater than S at any instant, work will be done upon the 
steamer and as such work cannot be lost it becomes converted 
into increased kinetic energy, and if F is less than S the kinetic 
energy of the steamer will decrease ; this is expressed in simple 
language by* saying that, if F is greater than >S' the speed of the 
vessel will increase, but if S is greater than F the speed will 




Fig. 28. 

decrease. The kinetic energy can be regarded as energy stored 
up for use in emergency ; if the effort is less than the resistance, 
the body gives up some of its kinetic energy to make up the 
difference between the work done by the effort and the resistance. 
When this difference in work is equal to the whole kinetic energy 
that the body possessed in the first place, the body will stop 
moving. 

This is the first time that we have dealt with the case in 
which a body may move in a direction opposite to that in which 
the resulting force upon it acts. When the direction of move- 
ment is opposite to that of the force we shall speak of the force 
as taking work from the body. 

Resistances are nearly always what may be called "induced" 
or "passive"; that is to say they disappear directly the body 
comes to rest. The resistance to the motion of a steamer increases 
very quickly with the speed and we soon get to a speed which 
may be regarded as the most economical. A slight increase of 



46 WORK, POWER AND ENERGY [ch. 

speed over this will require more coal and cost more money than 
the saving in time is worth. 

We have another similar example in racing motor cars. An 
ordinary 12 h.p. car can do 30 miles per hour but to get 80 miles 
an hour we have to increase the horse-power to something like 
80 or more. 

In nearly everything that engineers have to deal with, it is 
energy that they must try to use to the best advantage, because 
money is merely the token for energy. If the world's supply of 
coal, oil and other fuel gave out, it would take very few years 
before we should nearly all be starved to death. James Watt*s 
improvements of the steam-engine probably did more for the 
benefit of humanity than any scheme that human skill has 
devised, because it opened up vast fields for the use of the energy 
stored up in fuel. 

Graphical representation of Effort and Resistance. Suppose 
that the effort in moving a body from a point X to a point T 
varies in the manner indicated by the curve ABC, Fig. 29, and 
that the resistance varies in the maimer indicated by the curve 
DBF. Then if we take two points KL very close together on 
the base — so close that the effort F and resistance 8 may for all 
practical purposes be considered as constant over the length — 
the work done upon the body by the effort from ^ to L is equal 
to force X distance = F x KL = area of strip EGLK. Therefore 
as already shown the total work done on the body by the effort 
in moving from X to T is equal to the area ABGTX. 

Similarly the work taken by the resistance from the body in 
moving from ii" to L is equal to S x KL = area of strip HJLK ; so 
that the total amount of work taken from the body in moving 
from X to 2" is equal to the area DBFTX. 

Now the resultant work on the body is equal to work done 
by the effort — work taken away by the resistance 

= area ABCTX - area DBFTX 
= area ABD — area BFG. 

At any interniodiate point such as K the excess of work done 
by the effort over the work expended in overcoming the resistance 
is the difference between the areas XAEK and XDHK. 

Therefore between the points X and U the body increases in 
kinetic energy by the amount represented by the area ADB and 



Ill] 



WORK, POWER AND ENERGY 



47 



it then loses in going between U and T an amount of kinetic 
energy represented by the area BFC. We have not yet explained 
how the kinetic energy can be expressed in terms of the velocity 
but we have seen that the velocity is an indication of the kinetic 
energy ; consequently at the point U the body has the maximum 
amount of kinetic energy and therefore has the maximum velocity 
or speed. 



Qafn in Kinetic Energy 




Fig. 29. Work against Resistance. 

If the conditions were reversed so that the resistance were at 
first greater than the effort and less at the end, the body would 
be losing kinetic energy up to the point U\ U would then be 
the point of least kinetic energy and therefore of least velocity. 

We shall deal later with many problems concerned with 
kinetic energy ; our present aim is just to make clear the idea of 
work and energy and the fact that energy is never destroyed. 

Numerical Example of Effort and Resistance. A body is 
being urged fonvard by a constant force equal to 100 lbs. and over 
a distance of 120 feet the resistance increases uniformly from 30 lbs. 



4<9 



WORK, POWER AND ENERGY 



[CH. 



io 150 lbs. At what point will the body move with the greatest 
velocity ? How much kinetic energy will the body then have gained 
and how much will it have gained at the end of the 120 feet ? 

Referring to Fig. 30 ABC is the effort curve and DBF is the 
resistance curve. 

The point U gives the point of maximum velocity. The 
distance XU can be measured by drawing the diagram to scale, 
e.g. distances to a scale 1" = 20 feet and forces to a scale V = 50 lbs. ; 
one square inch of area would represent 20 x 50 = 1000 ft. -lbs. 
It will come to 70 feet. 



t 

lOOlb 



t 
30 

X 




Res'istancf 
Curve 



t 
150 



120 ft. 



Fiff. 30. 



By calculation we should proceed as follows. Draw DY 
horizontally as indicated in dotted lines. 

DV BV 
Then -^^, = r,^r because SF is parallel to FY, 
DY FY 

. DY.BV _ 120 (100 - 30) _ 120.70 _ 

• • ^^ = ~~FY- - ~ lW-T0"~' - "120 - ^^^^*- 

Gain in k.e. up to C7 = area of A ABD = \AB . AD 

= J. 70. 70 
-2450 ft. -lbs. 



Ill] WORK, POWER AND ENERGY 49 

Gain in k.e. up to 2^ = area of A ABD — area of A BFG 

= 2450-1.50.50 
= 2450 - 1250 
= 1200 ft.-lbs. 

Mean Effort. It is sometimes convenient to find the uniform 
effort which acting over the same distance will do the same 
amount of work as a variable one ; this is called the mean effort. 
Referring to Fig. 26 let F^^ be the mean effort ; then work done 
by F^ = Fjn X HJ. But the work done by the mean effort has 
to be equal to the work done by the variable effort. 

.-. F^x HJ = area HAKDJ, 
^ _ area HAKDJ 

Expressing this in general terms we have 

Area below effort curve 



Mean effort = 



Length of effort curve 



SUMMARY OF CHAPTER III. 

The work done by a force upon a body is measured by the product 
of the force and the distance through which the body moves in the 
direction of the force. The unit is the foot-pound. 

The work depends only on the initial and final positions of the 
body and not upon the path taken between the points. 

Power is the rate of doing work, i.e., the number of foot-pounds 
of work done per unit of time. 

One Horse-Power is equivalent to 33,000 foot-pounds per minute. 

Energy is the capacity for doing work and can exist in various 
forms which can be converted from one to the other. 

Mechanical energy can be divided into two kinds : kinetic energy 
(energy of motion) and potential energy (energy of position). 

The law of the conservation of energy states that while energy 
can be converted from one form into another, it can be neither 
created nor destroyed. 

If the force or effort be plotted against the distance, the result 

A. M. 4 



60 WORK, POWER AND ENERGY [oh. 

is called the effort curve and the work done up to any point is 
represented by the area of the effort curve up to that point. 

The work curve is the sum curve of the effort curve. 

In doing work upon a body against a resistance, the difference 
between the work done by the effort and the work done against the 
resistance goes in changing the kinetic energy of the body. 



EXERCISES. III. 

1. A chain 200 yards long and weighing 6 lbs. per ft. hangs 
vertically down a mine shaft. Find the work done in hauling it 
to the surface. 

2. If in the preceding a weight of | ton is attached to the end 
of the chain, find the total work done. Express each of the above 
results by a diagram. 

3. Find the horse-power of an engine which will lift the weight 
in Question 2, in 25 seconds. 

4. Find the horse-power required to pump 3000 gallons of 
water from a depth of 250 ft. in 10 minutes. 

5. How many cubic ft. of water would an engine working 
at 100 H.p. raise per min. from a depth of 25 fathoms? 

6. Find the work done in excavating a circular well 8 feet 
diameter, 45 feet deep, the weight of 1 cubic yard of earth being 
1 ton. Give answer in ft. -lbs. 

7. A horse drawing a cart at the rate of 2 miles per hoiu* exerts 
a tractive force of 156 lbs. weight. Find the work done in 1 minute. 

8. How many horse-power would be required to raise 2000 
cubic feet of water per hour from a mine whose depth is 180 fathoms ? 

9. Find the horse-power required to draw a train along a level 
at 45 miles per hour, whose weight is 250 tons, the resistances being 
taken at 15 lbs. wt. per ton. 

10. A cage with coals together weighing 10 cwt. is carried on 
the end of a wire rope weighing 10 lbs. per yard. Find the work 
done in ft. -lbs. in lifting it from the bottom of a mine 1500 ft. deep. 

11. The travel of the table of a planing machine which cuts 
both ways is 9 ft. If the resistance to be overcome while cutting 
be taken at 400 lbs. and the number of double strokes per hour 
be 80, find the h.p. absorbed in cutting. 



m] 



WORK, POWER AND ENERGY 



51 



12. When a prismatic column of stone, 20 ft. diameter outside, 
10 ft. diameter inside, 90 ft. high is being built, what actual work 
is done in lifting the stone from the ground ? One cubic ft. of stone 
weighs 125 lbs. 

13. What must be the effective h.p. of a locomotive wliich 
moves at the steady speed of 35 miles an hour on level rails the 
weight of the engine and train being 120 tons and the resistances 
16 lbs. per ton? What additional h.p. would be necessary if the 
rails were laid along a gradient of 1 in 112? 

14. Each of the two cylinders in a locomotive engine is 16" 
diameter and the length of crank is 1 ft. If the driving wheels 
make 105 revolutions per minute and the mean effective steam 
pressure is 85 lbs. per sq. in., what is the h.p.? 

15. A chain hanging vertically 520 ft. long weighing 20 lbs. 
per ft. is wound up. What work is done? 

16. A 10-ton hammer falls through a height of 6 feet and makes 
an impression on a mass of iron to the extent of 1 in. Find the mean 
statical pressure in tons which has been exerted on the mass of iron 
during the blow. 

17. A body weighing 1610 lbs. was lifted vertically by a rope, 
there being a damped spring balance to indicate the pulling force 
F lb. of the rope. When the body had been lifted x ft. from its 
position of rest, the pulling force was automatically recorded as 
follows : 



X 





11 


20 


34 


45 


56 


66 


76 


F 


4010 


3915 


3763 


3532 


3366 


3208 


3100 


3007 



Find approximately the work done on the body when it has risen 
70 ft. How much of this is stored as potential energj'^ and how 
much as kinetic energy ? 



4—2 



J 



CHAPTER IV 

MACHINES AND EFFICIENCY 

A machine may be described as an appliance for receiving 
energy from some outside source and delivering it in some 
more convenient form for doing work. Almost the simplest 
possible form of machine is the lever, which in the form of the 
crow-bar is used for lifting heavy packing cases. A man unaided 
cannot move the case; that is to say he cannot exert a force 
sufficiently great to lift it. He possesses quite enough energy 
to do so, but he can exert only a comparatively small force to 
move a body, although he can continue to exert it over a long 
distance ; whereas for lifting the case he requires to exert a large 
force over a short distance ; the crow-bar enables him to do this. 

We have another every day illustration in the use of two and 
three speed gears in bicycles. When the cyclist comes to a hill, 
he puts in the low gear. This does not give him any more 
energy, in fact it makes him lose a little more than usual on 
account of the extra complication of the mechanism, but it 
enables him to use his energy more conveniently. He goes more 
slowly up the hill but does not have to push so hard and he 
finds that the result is a gain in comfort. 

Wheel and Axle. We have a very simple form of machine 
in the wheel and axle shown in Fig. 30 a and we will show that 
we can get the same result by considering the work done as 
by considering the moments of the forces acting. In many 
problems a consideration of work done gives the quickest results. 

A weight W to be lifted is connected by a rope or chain to 
an axle B of radius r which is supported in bearings and carries 
a wheel C of radius R to which the effort F is appUed. 

By moments we should have Wr = F . R, 

IP ^'" 



CH. IV] 



MACHINES AND EFFICIENCY 



63 



Now suppose that the axle makes one revolution. 

The weight W moves up a distance 27rr and the work done 
upon the weight is therefore W . 27rr. 
The rope to which F is applied moves 
downwards a distance 27tR so that the 
effort does an amount of work equal to 
F . 27rR. If the axle runs freely, these 
two amounts of work must be equal. 

.-. F . 2ttR = W . 27rr, 

or ^ = ^, 

as before. 




Crow-bar. Referring to Fig. 31, the 
full lines indicate the position of the i^„ 3q^^ 

crow-bar after the case has been raised 

while the dotted lines indicate the position before raising. F is 
the effort that the man can exert upon the end A of the lever 
and S is the resistance which the lever exerts upon the case at 
the point B. If O is the pivot or fulcrum of the lever and the 
perpendicular distances from G to the lines of action of F and 
S are respectively x and y, we have seen already that by the 
principle of moments 

F X x = S X y (1). 

If therefore F = 100 lbs. ; a; = 30 inches and y = 2 inches, 

^ Fx 100 X .30 __„„ 
jS = — = ^ = 1500 lbs. 

y 2 



(/ 



Mechanical Advantage. In a machine for converting one 

3sistan( 
Effort 



form of mechanical energy into another the ratio -^,v — r— is 



called the mechanical advantage. 

In our particular case above we have 

Mechanical advantage =• ^ = ~rKrr =15. 

It will be noted that the " arm " a: of the effort F in the position 
shown in full lines is appreciably larger than in the original 
position shown in dotted lines and that the arm "y" of the 
resistance S does not change appreciably; this shows that the 



54 



MACHINES AND EFFICIENCY 



[CH. 



mechanical advantage of this particular machine increases as the 
weight is lifted, so that the effort F will gradually diminish. 

Now let us suppose that the distance A^A^ is so small that 
7^ and S are for all practical purposes constant during the lifting 
action. Then we have 

Work done by effort = F x h. 
Work done by resistance = S x z. 

If no work is wasted by frictional forces at the fulcrum or 
pivot, these two amounts of work must be equal or 

Fh = Sz (2). 



^ 




(3). 



Fig. 31. Crow-bar. 

We have already shown by (1) that 

Fx = Sy. 
Therefore dividing we get 

h _ z 

x~ y 

We will now try and prove that this should be the case ; the 
student should try to verify this by drawing to scale and measuring 
to length. This is particularly desirable because the proof is 
rather long. We have already explained that the distance A^A^^ 
is very small ; therefore the angle a will be small ; so small that 
the line A^A.^ will not be appreciably different in length from an 
arc with centre (7. 



IV] 



MACHINES AND EFFICIENCY 



55 



Now angle = arc 



a 



-r radius ; 



AoC 



and also a = -vl^r^ = - 



A.1A2 
A^C 



Again the angle. ^1^2^ is practically a right angle; 
which is 90° — GA2C is equal to ft; and 



(4). 



GCA 



.*. sin ft =- 



h 



A1A2 
h jf' 



and also = 
A1A2 



OA, 
GA, 



X 



that is = ^^-r- 



^■9^.2 KjJxt 



or 



A2G 



- = - (from (4)). 



This is the result that we attempted to prove. We see there- 
fore that the principle of moments gives us the same result as 
the principle of work; in some problems it is more convenient 
to use moments and in others it is more convenient to consider 
the idea of work. 

Efficiency of Machines. It is of considerable help in the 
understanding of the mechanical principles of machines to 
imagine a machine to be a kind of box, as indicated in Fig. 32, 

I Energy Input 




Energy Output 

Id ^^O 

O 

Fig. 32. 

provided with an inlet / into which a certain amount of energy 
Ej is put in in a given time and with an outlet through which 
issues an amount Eq of energy in the same time. Now in every 
machine a certain amount E^ oi energy is lost or wasted. There- 
fore we may write Inlet Energy = Outlet Energy + Waste Energy. 

Or in symbols Ei = Eo-\- E^ (5). 



56 MACHINES AND EFFICIENCY [ch. 

Now the quantity -~- is called the Efficiency of the machine. 

Expressing this in words we should say that the efficiency of 
a machine is the ratio of the energy that it gives out to the energy that 
it receives. 

Suppose, for instance, that a certain machine receives 120,000 

ft. -lbs. of energy in a certain time and that during the same time 

it gives out 97,000 ft.-lbs. 

97 000 
Then Efficiency = e = , n^-r^KK = '808 nearly. 

It is the usual practice to express efficiencies as so much per 
cent., i.e., to multiply the actual efficiency by 100. In our case, 
therefore, we should then say that the efficiency is 80-8 per cent. 

Again since the energy wasted E^r^ Ej — Eq we have 

|":=l-f-»=(l-c); 

:. energy wasted = (1 — e) x energy input. 
The highest possible efficiency that a machine can have is 
1 or 100 per cent, but most machines have an efficiency consider- 
ably less than this, the simpler machines generally having a higher 
efficiency than the more complicated ones. The principal aim 
that an engineer has in designing machines is to make the efficiency 
as high as possible, that is to make the energy wasted as small as 
possible. 

Velocity ratio of Machines. In a machine in which the 
effort and resistance are constant in direction, the quantity 

Distance moved at the effort in a given time 
Distance moved at the resistance in the same time 

is called the velocity ratio. 

In the example shown in Fig. 31 we should have 

Velocity ratio = - = F, . 

In this case we showed that the mechanical advantage if there 
was no loss was 

'F~ y~ z' 

Therefore when there is no loss of energy mechanical advantage 
= velocity ratio. 



IV] MACHINES AND EFFICIENCY 67 

When, however, some energy is lost we have to modify this 

result because, although the velocity ratio is fixed by the actual 

sizes of the elements forming the machine, the mechanical 

advantage depends upon the amount of energy that is wasted. 

As a more general rule therefore we say that 

Mechanical advantage = velocity ratio x efficiency, 

-, . ,«- 1 . , 1 . Resistance 

and smce Mechanical advantage = — ^^rs — i — > 

° Effort 

we may say that 

Resistance = effort x velocity ratio x efficiency, 

i.e. S=F.Vr^e. 

In the case of the crow-bar that we have examined the velocity 
ratio is 15. Now suppose that the pivot is rough so that energy 
is absorbed in moving it and suppose that 3 % of the energy 
input is wasted, then efficiency = e = 1 — yf^ = -97. 
Then we should have 

/8= 100 X 15 X -97 = 1455 lbs., 
or if we require to find the value of F for S = 1500 we have 

1500 = jP X 15 X 97; 
.-. F = 103 lbs. 

Now the effort that would be required in a perfect machine, 
in which no energy is wasted and whose efficiency is therefore 1, 
we shall call the ideal effort. 

Resistance 



Now Actual effort = 

and Ideal effort = 



Mechanical advantage * 
Resistance 



Velocity ratio ' 
Ideal effort Resistance Resistance 



Actual effort Velocity ratio ' Mechanical advantage 
Mechanical advantage 



= Efficiency, 

Trlipa.l P:ffnrh 

i.e. Actual effort = 



Efficiency 

Some simple Machines. The inclined plane. The inchned 
plane is one of the most ancient forms of machine and is one 
of the simplest. Suppose, for instance, that we wish to raise a 



58 



MACHINES AND EFFICIENCY 



[CH. 



V 



i 



body such as a truck up to a point B. It is too heavy to lift, but 
by running an inclined plane AB from the ground level to the 
point we can push the body slowly up. 

(a) Effort parallel to the plane. 

If F (Fig. 33) is the effort or force pushing the body up the 
plane, and acting parallel to it, the work done by the effort in 
moving the body up is jF* x AB. The body has to be raised a 
vertical distance BG ; this is the distance moved in the direction 
of its weight, representing an amount of work equal to TF x BC. 




F_,3 




Fig. 33. Inclined Plane. Effort x>arallel to Plane. 

If the velocity is constant between A and B there is no change 
in kinetic energy and if also there is no energy wasted we have 

Work done by effort F = Work done against weight W ; 
:. F X AB = W X BC, 

or 'w ^ ^RP "^ Mechanical advantage. 

., TT I .^ X- Distance moved in direction of F 

Also Velocity ratio = ^.- ^— — 3% — 7^— .-^5^ 

Distance moved m direction of W 

_ AB 

~ BC 

We can also find a relation between F «nnd W by considering 
the forces acting upon the body. The third force is the reaction 



IV] MACHINES AND EFFICIENCY 59 

R between the truck and the plane; if there is no friction this 
reaction will be at right angles to the plane. Therefore by 
drawing lines parallel to the forces to a convenient scale we have 
a triangle of forces or vector figxire 1, 2, 3. y 

It will be noted that each side of this triangle is at right 
angles to a side of the triangle ABC ; therefore the two triangles 
are similar, therefore 

W^2^^AB 
F ~ 2,1 ~ BO' 

1? W.BC ... 

^=-AB- (^)- 

Ai R 1,3 AC 

^*^° W ~ 2/ 3 ~ AB ' 

•••^=^1S (2). 

We can use the language of trigonometry to express these 
results as follows : 

^^=sm^; 

/. E= Wsine (3), 

= cos^; /. JS=Tfco3^ (4). 

We also note that 

4. c BC 
tan 6 = -j-pf . 
AG 

/ 
Numerical Example. What force is necessary to push a truck 

weighing 15 tons up a gradient of 1 in 10 ? 

This means that BC is 1 when AC is 10. 

Then since AB^ = BC^ + AC\ 

AB = jr+ 100 = VlOl = 1005 nearly; 

„ 15 X 1 , ^^ ^ 

•*• ^ = lAAg = 1*49 tons. 
lU'Uo — — — — — 

In practice F will always be more than this because there is 
always some energy wasted, the principal cause of waste being 
called friction. The above value of ^ is the Ideal effort. 



AC 
AB 



60 



MACHINES AND EFFICIENCY 



[CH. 



Experiment. A very simple piece of apparatus can be rigged up for 
the experimental verification of the laws of the inclined plane. It is constructed 
by fixing a board C (Fig. 34) by a hinge D to one end of a board A. A vertical 
board iJ is fixed at the end B and is provided with a slot through which passes 
a shouldered pin provided with a fly-nut so arranged as indicated in the 
detailed figure that the board C rests on the projecting pin. A scale *S is carried 
by the vertical board E and a pulley P is fixed in the free end of the board C. 




Fig. 34 

A light string is attached to a truck Q in which is placed a weight, the combined 
weight of the truck and weight being equal to W ; the string passes over the 
pulley P and through a slot in the end of the board C and has its other end 
attached to a spring balance J. 

We then measure carefully the distance x from the hinge to the edge of 
the scale and write it at some convenient place on the apparatus. 

The pin is then set to a certain value of y, the fly-nut tightened up, and the 
reading on the spring balance is noted. 

If we wish to save time in our calculations by using trigonometrical methods 
we next proceed to calculate the value of the angle 6 for various values of y 
and draw a calibration diagram. 

Suppose for instance that a; = 30 inches. 

When 2/ = 5 inches we have tan d=^^~'\Q&l ; from trigonometrical tables 
we find that ^ = 9'5 degrees. 

When y = 10 inches we have tan ^ =^ = -3333, and we get by tables d = 185 
degrees about. 

Similarly we get y = 15 ^ = 26*6 degrees. 

20 =337 

25 =39-8 

30 =45 



IV] MACHINES AND EFFICIENCY 61 

We then by plotting obtain the calibration diagram shown in Fig. 36. 

45 
40 
35 
80 
25 
SO 
15 

a 
f-i 

O 

d 






5 

y in inches 



10 



15 



20 



25 



30 



Fig. 35. 



From this diagram we can read off at once the angle 6 for any value of y 
and can calculate the theoretical values of F. The results can then be tabulated 
as follows: 



Wlha. 


y inches 


Spring balance 
reading 
=F lbs. 


d 

[from 
diagram] 


Ideal value 

of ^ 

= W8md lbs. 













If we wish to avoid trigonometry we note that the quantity corresponding 
ix) AB in Fig. 33 = s/3e^Ty^; 



Theoretical value oi F = 



s!x* 



W^y 
rj^+y^' 

The above gives what is called a static test. To get an approximation 
to a dynamic or running test, take away the spring balance and replace it by 
a scale pan. Let the truck rest against the stop and weight the pan untU 
the truck moves up slowly without gaining in speed; then the combined 
weight of scale pan and weights gives the effort F required to move the truck 
up without increasing its speed and therefore its kinetic energy. Now take 
the weights off slowly until the truck begins to run back without increasing 



62 



MACHINES AND EFFICIENCY 



[oh. 



y 



its speed ; the resulting weight is the effort that the truck can exert in moving 
down. We can then tabulate conveniently as follows: 



IF lbs. 


y inches 


e 

[from 
diagram] 


Ideal value 

of i^ 
= H^sin^lbs. 


Observed value of F in lbs. 


Rising 


Fallmg 















(6) Effort parallel to the ground. Now let the force F act 
horizontally as indicated in Fig. 36. When the body has moved 





NrW 



Fig. 36. Inclined Plane. Effort horizontal. 

from A to By it has gone in the direction of the effort F a distance 
AC and in the direction of the weight W a distance CB. If 
therefore no energy is wasted 

Work done by effort F = Work done against weight W 

or FxAG=Wx BC, 

i.e. "rf = D/i = Mechanical advantage; 

.-. ^=TF.^ = TFtan^. 
AC 

Considering the subject from the standpoint of equilibrium 
of forces we have 1, 2, 3 as the vector figure or triangle of forces. 
It will be noted that this triangle is the same as the triangle 



IV] 



MACHINES AND EFFICIENCY 



63 



.-. F 



ABC drawn to a smaller scale and turned through ninety 
degrees. 

Therefore the triangles are similar or 

W~2,S~AC'' 

W.BC , , 

= j-^ — , as before. 
AG 

TF ~ 2, 3 AC 

The dotted line on the vector figure shows the value of F in 
case (a). 

The Screw. The screw is a form of inclined plane, and may 
be considered as an inclined plane wrapped round a cylinder. 

S 



Also 




Fig. 37. The Screw. 

It is formed by cutting a groove in the cylinder leaving a pro- 
jection or thread which may be either of triangular or square 
form. In Fig. 37 is shown a square thread. It is engaged by 
a nut N having a corresponding thread in it. If the nut is fixed 
and the screw shown is turned in a clockwise direction the screw 
moves further down into the nut; such a screw is called right- 
handed. If on rotating the screw in a clockwise direction it 
moved upwards out of the nut, it would be called left-handed. 
The distance moved into or out of the nut in one turn is called 
the pitch of the screw; the pitch might also be defined as the 



64 MACHINES AND EFFICIENCY [c5H. 

distance parallel to the axis of the screw between corresponding 
edges of two successive threads. 
V If we consider one thread of the screw as unwound until it 
is all in one plane we should get the inclined plane ABC. 

Now suppose that the screw supports a weight W and that 
it is prevented from rotating ; if the nut is turned in a clockwise 
direction by means of a spanner T, a force F^ being applied at 
a distance x from the centre of the screw, the weight will be 
lifted. 

Then if F^ is altered in direction so that it is always at right 
angles to the spanner as the latter rotates, the distance moved in 
the direction of F^ in one revolution will be equal to the circum- 
ference of a circle of radius x, i.e. 27tx, so that the worJ^ done by 
Fi is equal to F^ x 27tx. In this one revolution of the nut the 
weight will be lifted by an amount p so that the work done on 
the weight will be ^qual to W x p; 

.*. we have F^ x 27tx = W x p 

We may also consider the problem as follows; the force F^^ 
Sit the end of the spanner is equivalent to a force F at the screw 
thread. 

By taking moments about 0, the axis of the screw, we have 

Fixx = —^. 
By our previous treatment of the inclined plane we have 

F.^Wt^nO^—^; 

„ Fd Wp d 

..i'lXa;- 2 - ^^ -2 

_Wp 

Wp 
:. T\ -= , as before. 

This is of course the ideal ellort; in practice it will be more 
on account of the energy wasted due to the friction between the 
nut and the thread and between the nut and the fixed surface U. 



IV] 



MACHINES AND EFFICIENCY 



65 



Screw-jack. A very common form of machine employing the 
screw is the "screw-jack" which is used for lifting heavy bodies 
through short distances and is used largely for lifting motor cars 
at one side in order to remove the wheel. 

It consists of a screw A (Fig. 38) working in a nut formed in the 
top of a base B, the end of the screw ending in a knob portion pro- 
vided with *' tommy-holes" D, the extreme end being turned to 
a smaller diameter and carrying a thrust cap E provided with 




Fig. 38. Screw-Jaok. 

ridges to give a good grip. A hand-lever C is passed through 
one of the tommy-holes and is pushed round in the direction of 
the arrow ; when the lever has been given about a quarter turn, 
it is put through the next hole and pushed round further thus 
slowly raising the article to be lifted, the nut being fixed. 

Numerical Example. A screw-jack has the screw of V pitch 
and the lever is 15 inches in length from the centre of the screw to 
the point at which it is grasped. What force must be exerted on 
the lever to lift a load of 2 tons if the efficiency of the machine is 40 % ? 

Suppose that the lever makes one complete turn. 

Distance moved by effort == 27r x 15 inches. 

A. M. 5 



/■ 



66 MACHINES AND EFFICIENCY [oh. 

Distance moved by resistance or weight = pitch of screw 
= J inch. 

.*. Velocity ratio = -^ — = 607r = 188-5; 

2 

' . T J 1 i» J. Resistance 

.*. Ideal effort = ^f-s — n 1^ 

Velocity ratio 

2 X 2240 



= 23-8 lbs. 
.*. Actual effort 



188-6 
Ideal effort 23-8 



Efficiency -4 
= 59 5 lbs. 

Reversing Machines. If the efficiency of a machine be 
sufficiently great it will, if allowed, reverse, that is the resistance 
acting as an effort will be able to make the macliine run back- 
ward, but if the efficiency be less than 50 % this cannot happen. 

For let the input, the output, and the waste energy, when the 
machine is acting direct, be respectively Ej, Eq^ and E^, then 

Ej= Eq-\- EjfT' 

Now let the resistance act as an effort and do work Eq, the 
body moving through the same distance as before ; the amount 
of waste energy is again Ej^ and the balance Eq - E^ will be 
available as output at what was originally the effort end of the 
machine. If ^o is greater than JS7,^, that is if the efficiency is 
greater than 50 %, there will be some work delivered, but if 
E^ is greater than Eg the resistance will not even be able to 
overcome the wasteful forces, that is the machine cannot run 
back unaided. 

This general explanation may be a little difficult to follpw at 
first but will probably be made clear by the following numerical 
illustration. 

Suppose that we have a machine with velocity ratio 10 and 
efficiency *4 and let the resistance be 100 lbs. 

Also let the part of the machine at which the effort is applied 
move through 10 feet in the direction of the effort; then the 
resistance end moves through 1 foot in the direction of the 
resistance. 

Then Input energy E = 25 x 10 = 250 ft. -lbs., 
Output energy Eq = 100 x I = 100 ft.-lbs. ; 
.-. Waste energy E^ = 250 - 100 = 150 ft.-lbs. 



IV] 



MACHINES AND EFFICIENCY 



67 



Now let us reverse the conditions and allow 100 lbs. to act as 
an effort through 1 foot if it can ; it would do 100 ft.-lbs. of work 
which is not sufficient to supply the 150 ft.-lbs. of waste energy 
so that 100 lbs. will not be sufficient to reverse the machine. 

A machine that will not reverse is called self-sustaining. In 
some machines this is a convenience ; for instance in the case of 
the screw-jack previously described. In such cases we have to 
pay for the convenience by low efficiency. 

Pulley Tackle. The various forms of pulley tackle are 
examples of simple forms of machine. 





W 



^ 



Fig. 39. Pulley Tackle. 

Fig. 39 (a) shows one form. It consists of two blocks A, B 
each consisting of two pulleys of equal size. The rope is fixed 
to an eye C in the upper block and then passes over one pulley 
in the lower block; then over one of the pulleys in the upper 
block ; then over the other pulley in the lower block and finally 
over the remaining pulley in the upper block Fig. 39 (6) shows 
the arrangement diagrammatically, the two pulleys in each 

6—2 



/ 



68 MACHINES AND EFFICIENCY [ch. 

block being of slightly different diameters to show more oleaxly 
the manner in which the rope passes over them. 

Now suppose that the rope at the end F is moved downwards 
one inch, the lower block will then move upwards J inch because 
there are four ropes that have to move up by the same amount 
and the total amount of upward movement must be equal to 
the downward movement at the end F because the rope is 
continuous. 

If therefore there is no loss of energy we shall have 

Weston's Differential Pulley Block. This block consists of 
two specially grooved pulleys, A, B, Fig. 40 (a), of slightly different 
diameters cast in one piece and secured to a strong upper support. 
The grooves are formed with flat portions to engage the chain in 
the manner of teeth. The weight W is secured to a second 
similarly-grooved pulley C. An endless chain F passes over the 
larger pulley A; then over the pulley C; and then over the 
smaller pulley B, as shown, the effort F being applied to the chain 
that comes over the larger pulley. Now suppose that the larger 
pulley A is of effective diameter D inches and that the smaller 
is of effective diameter d inches. 

Guides are provided for the chain. 

Now suppose that the chain is pulled so that the upper pulleys 
make one complete revolution. The amomit of chain rolling off 
on the left, coming from the pulley B, will be rrd inches, and 
a length of the chain equal to nD inches will roll on on the right, 
so that the chain as a whole rolls on a distance equal to (ttD — wd) 
inches. 

The weight will move up half this distance or ^ ^inches. 

The reason for this half requires some further explanation; 
we will explain it by considering a rope or chain passing over a 
single pulley Q, Fig. 40 (6), and fixed at one end to a point P. 
Now suppose that the free end is moved from the position X to 
the position X' ; the pulley moves up to the position shown in 
dotted lines. The rope or chain may be considered aa made up 
of three lengths ; the piece FM on the left ; the piece M'N 



IV] 



MACHINES AND EFFICIENCY 



69 



encircling the pulley and the piece NX on the right. In the 
raised position the pieces are PM\ M'N' and N'X\ 




P 



/SX 



vlv 



Q 



(b) 



Kg. 40. Weston's Pulley Block. 

Therefore total length before movement = PM + MN -h NX. 
Therefore total length after movement = PM' + M'N' + N'X'. 
Now these two lengths must be the same and clearly 

MN=:M'N'i 
:, PM + NX = PM' + N'X', 
i.e. PM' + M'M + NN' + N'X = PM' + N'X + XX', 
or M'M + NN' = XX' ; 

but clearly MM' = NN' = distance moved up by Q, 

XX' 



or 



M'M =■ 



2 * 



70 MACHINES AND EFFICIENCY [ch. 

i.e. distance moved up by pulley = J distance moved up by rope 
or chain. 

Summarising our results we have : 

Distance moved at effort = itD. 



Distance moved at weight 
.*. Velocity ratio = 



2 

ttD 



ttD — Trd 



2~ 
2D 
D-d 



(1). 



.-. Ideal effort= i?'^ — ^"^^ (2). 

d is usually made nearly equal to D to make the ideal efforts as 
small as possible. 

Numerical Example. In a Weston Pulley Block the larger 

pulley has 12 teeth and the smaller has 11 teeth. If the efficiency 

is 60 per cent., what load will be raised by an effort of 20 lbs. ? 

2D 
In this case the velocity ratio fr- -j will be equal to 

li^=24; 
12-11 ' 

W 
:. Ideal effort = ht » 
24 

K 4. i rt 4. Wea^ effort W «^ 

Actual effort = -^^ . = sr -^ '60 ; 

Efficiency 24 

W 

.-.20 = 



24 X -60' 
Tr=24x -60 X 20 lbs. 

= 288 lbs. 

Actual Performance of Machines. We have already stated 
that in practice machines are never ideal and that some energy 
is always wasted. If we regard the actual effort as the sum of 
ideal effort and waste effort we shall find that in actual tests the 
waste effort is almost constant but increases slightly as the load 
or resistance increases. 

The usual procedure in testing a simple machine is to first 



IV] 



MACHINES AND EFFICIENCY 



71 



find what effort can be exerted, when there is no load on the 
machine, before the point at which the effort is applied will move 
slowly without increasing in speed. This initial effort is the 
initial waste effort and spends itself in Ufting the dead weight 
of the machine itself and in overcoming the friction or "sticki- 
ness" of the various parts. 

Various loads are then put on the machine and the effort 
necessary to lift each slowly at the same speed is noted carefully. 




A 



50 



100 150 200 250 M 300 350 

Load or resistance in lbs. 
Fig. 41. 



400 - .450 500 



To minimise errors it is a good plan to increase the loads gradually 
in, say, 10 steps, i.e. 50 lbs. at a time for a maximum load of 
500 lbs., until the maximum load is reached and then decrease 
the loads gradually by the same amount, the mean of the actual 
efforts "ascending" and "descending" being taken as the final 
values. The mean actual efforts are then plotted against the 
loads as indicated in Fig. 41 and the resulting curve will usually 
give a straight line CD. 

On the same base are then plotted the values of the Ideal 
Efforts. This will give a straight Une AG, given by the relation 

AB 

AG = — , where Vj. is the velocity ratio. 



72 



MACHINES AND EFFICIENCY 



[CH. 



Now since Efficiency = -r— - — -, — ^--r we can obtain values 

Actual effort 

from which we can plot to a convenient scale the efficiency curve 

AHJ. 

For any load, say ALy we have 

c = Efficiency = ^j- . 

It is preferable to take the values of the actual effort from the 
line CD, instead of from the observed values, because errors of 
observation are smoothed out by the curve. 

Experiment upon Weston Pulley Block. Take for example the 
Weston Pulley Block described on p. 69, and suppose that the loads are 
increased 60 lbs. at a time; the results thus obtained may be tabulated as 
follows : 



Resistance or 
Load = W lbs. 


Ideal Effort 
=r,lbs.=^ 


Mean Actual 
Effort rri" lbs. 


Efficiency 
_ Ideal Effort 






Actual Effort 








60 





00 


208 


8-9 


•234 


100 


417 


110 


•379 


ISO 


6-26 


131 


•477 


200 


8-33 


160 


•521 


200 


10-42 


18-1 


•576 


300 


12-60 


20-9 


-698 


350 


14-68 


22-9 


•637 


400 


16-67 


26-5 


•664 


460 


1875 


28-2 


-672 


000 


20-83 


301 


•692 



As an illustration of the suggestion of taking the values of the actual effort, 
for calculating the efficiency, from the curve CD instead of from the actual 
observed values we will take the case of TF = 360 lbs. The observed value 
of i^ is 22*9 lbs. but the value obtained from the diagram is 23*1 lbs. ; this 
value was used in calculating the efficiency. 

Experiments upon a bicycle gear. The following tests upon a bicycle 
provided with a two-speed gear were made in order to find the efficiency of 
the chain drive when driving " solid," and when driving through the speed 
gear, and can be made in a similar manner upon any bicycle. In order to 
make the nature of the experiment more clear we will first give a brief explana- 
tion of the action of such two-speed gears. Referring to Fig. 42 which shows 
diagrammatically the arrangement commonly adopted, the chain wheel of 
the back wheel of the bicycle is connected to an internally toothed wheel B 
with which engage a number of toothed wheels of pinions C — usually called 
" planet pinione " — which are carried by a cage D fixed to the hub of the 



IV] 



MACHINES AND EFFICIENCY 



73 



back wheel. The planet pinions C engage also a wheel A — asnally called 
the sun-wheel — which is capable of being fixed to the frame of the bicycle 
or of rotating freely. In the normal gear, the wheel A is free to rotate and 
a clutch or mechanical locking device locks the ring B to the cage D thus giving 
what is called a " solid drive." To put in the low gear the clutch between 
the ring and cage is released and the wheel A is fixed. The pinions C carried 
by cage D then have to roll simultaneously upon the fixed pinion A and the 
ring B and the cage is thus forced to go more slowly than the ring; this means 
that to drive the wheel, which is connected to the cage D, at a given speed the 
ring B and therefore the pedals must be rotated more quickly. In other 




Fig. 42. Bicycle two-speed gear. 



words the gear is lowered. Although we do not intend to explain the derivation 
of the formula at the present stage, the reduced gear can be calculated as 
follows : 

Let ^^=the number of teeth on the wheel A^ 

^^ssthe number of teeth on the annular wheel JB. 



Then 



Reduced gear = 



■X Normal gear. 



Now the " normal gear " of a bicycle is the diameter in inches of the equiva- 
lent direct driven single wheel. It is obtained by the rule: 

Normal gear in inches 

_ Diameter of back whe el in inches x Teeth on front chain wheel 
~ Teeth on small chain wheel 

In the case of the bicycle under consideration we get 

Normal gear in inches = — ^tt- = 80*89. 



18 



We also have Na = 21 and iV^£=69. 



T 80-89x69 80-89x69 -- , . , 
.*. Low gear= „^ ^, = — -^ =68*1 mchcs. 



69-i-27 



96 



74 



MACHINES AND EFFICIENCY 



[CH. 



Method of testing. To test the efficiency of the bicycle gearing the bicycle 
is suspended in the manner indicated in Fig. 43 and a scale pan B is connected 
by a string to the tyre. The cranks are placed in horizontal position and a 
scale pan A is suspended from one of the pedals so that when loaded it will 
tend to lift the scale pan B. In order to make the test as accurate as possible 
care must be taken that the back wheel is brought by the aid of the free-wheel 
clutch to the " balanced " position before the scale pan B is fixed to it. The 
tyre- valve is the principal cause of the lack of balance of a bicycle wheel and 
if a bicycle be lifted off the ground the wheel will start swinging due to the 
lack of balance and the position of the wheel at wliich it ultimately comes 
to rest is the one that is referred to above as the " balanced " position. 

In making the test, the scale pan B is weighted and placed upon its stop D 
and the scale pan A is then loaded carefully until it begins to fall slowly on 




Fig. 43. 

to its stop C ; the load is th en carefully taken off until it begins to rise slowly 
and the scale pan B falls on to its stop D. 

The following results were obtained in an actual test. 

Weight acting on tyre (including scale pan) = 2 08 lbs. 

Normal gear'. 

Weight acting on pedal to lift scale pan 5 = 12 25 lbs. 

Weight acting on pedal to allow scale pan B to lower = ir76 lbs. 
Low gear: 

Weight acting on pedal to lift scale pan JB = 9"35 lbs. 

Weight acting on pedal to allow scale pan B to lower = 7 75 lbs. 

We will now work out the velocity ratio. 
The cranks are 7 inches long. 

In one revolution of the crank, the distance moved by the centre of the 
[tedal = 2n- X 7 inches. The wheel moves through a distance = tt x gear. 

7rxl4 __J_ 
6-78* 

Low gear velocity ratio =:- " ^ '''' 



,'. Normal gear velocity ratio = 



ttX 80-89 

7rxl4 

TX581 415* 



IV] MACHINES AND EFFICIENCY 75 

Normal gear efficiencies. 

Ideal effort to lift scale pan jB= resistance x velocity ratio = 2 08 x5 78 

= 12 02 lbs. 
_„ . Ideal effort 1202 

•'• ^®"'^°°y= Actual effort = 12^5 

= •980 or 98-0 percent. 

1 1 .«(» 

Ideal resistance lifted by scale pan B = — _-. =2 03; 

0'7o 

.*. Reversed efficiency = ^:^ = '974 

=97*4 per cent. 

Low gear efficiencies. 

Ideal effort to lift scale pan £ = 208 x4'15 

= 8-63 lbs. 

_„ . ' Ideal effort 8-63 

•••^®'^'°°y= Actual effort = 9^ 

= •923 

=92 3 per cent. 

7'75 
Ideal resistance lifted by scale pan B = t^;^ =1'8Q7 ; 

1 *Sfi7 

/. Reversed efficiency = -^^^7^ =*897 

= 89*7 per cent. 

From the above we see that the efficiency at normal gear is 98*0 and at 

low gear 92*3; the two-speed gear therefore causes an additional loss of 

5*7 
980 -92-3 =5-7 in 98*0 or about ^^^x 100 = 58 per cent. 

Work done on rotating bodies. In the cases that we have 
considered up to the present we have dealt only with bodies 
which are moved in a straight line. In a larger number of cases 
in engineering practice, however, we have to deal with rotating 
bodies. Take for example the case of a pulley Aj Fig. 44, which 
is being rotated by a belt or chain B. There is a tension T^ lbs. 
on the tight side of the belt and a tension T^ lbs. on the slack 
side. 

Suppose that the pulley makes one revolution and that the 
belt does not slip and that the radius to the centre of the belt 
is r feet; the circumference of the pulley then moves the same 
distance as the belt, i.e. a distance equal to 27rr in the direction 
of the effort and resistance. 

The work done on the pulley by the tension Tj^ 

-= 27rrTi. 



76 



MACHINES AND EFFICIENCY 



[cu. 



Work taken from the pulley by the tension T^ 

therefore resulting work done on the pulley in one revolution 

- 27TrT^ - 2nrT^ 

= ^ = 27rr (Ti - Ta) (1). 

This represents the work done on the machine which the 
pulley drives. As a rough approximation we may take the tension 
on the tight side equal to twice that on the slack side. 



Slack side 




.Tight sido 
of Belt 



Some people derive this result as follows : 
Take moments about the centre of the shaft upon which 
the pulley is mounted. Then resultant moment 

= T^r - T^r = {T^ - T^) r. 

This resultant moment is called the torque. This gives rise to 
the following general rule: 

Work done per revolution inft.-lbs. = 27r x torque in lbs. -ft. . . (2). 

Now suppose that the pulley makes N revolutions in one 
minute. 

Then the work done on the pulley in one minute 

':-EN = 27TrN {Ti - T^) ; 

„ ^ Work per min. 27rriV (T^ - To) 
.-. Horse-Power = ^33^^00— = — MTWO" ' ' (3)- 

Numerical Example. What Jwrse-power is transmitted to a 
pulley rotating at a speed of 120 revolutions per minute if the 



IV] 



MACHINES AND EFFICIENCY 



77 



tension on the tight side is 150 lbs. and on the slack side is 75 lbs., 
the diameter of the pulley being 18 inches ? 

In this case T^-T^= 150 - 75 = 75 lbs. ; 

r= ^= '75 foot; 
.*. Work done per minute = 27r x '75 x 120 x 75 ; 

27r X -75 X 120 X 75 



.'. Horse-Power = 



1-29. 



33,000 



Indicated and Brake Horse-Power of Engines. In testing 
steam, gas, oil and similar engines it is usual to measure what 



Atea - Work done 
I in one rerolution 
^ -of shaft. 






t 






= 


r 




s 




■ ■ ■■- 




3 


F--^- 







-> Out stroke 



Instroke 



Fig. 45. Indicated Horse-Power. 

are called the "Indicated Horse-Power" (i.h.p.) and the ''Brake 
Horse-Power" (b.h.p.). 

The indicated horse-power is in a sense a measure of the power 
input and is calculated from diagrams drawn by an instrument 
called the "indicator" which automatically indicates graphically 
as a diagram the pressure of the steam or gas in the engine 
cylinder at the various points of the stroke. This diagram in 
the case of a steam-engine is somewhat as indicated in Fig. 45. 
The total pressure acting upon the piston is the effort so that the 
indicator diagram draws for us the effort curve and we have 



78 MACHINES AND EFFICIENCY [oh. 

shown already that the area under the effort curve represents 
the work done. 

On the outstroke of the piston the work done is represented 
by the area AG J KB and on the instroke the work taken from the 
piston in bringing it back is represented by the area BKLGA ; 
the difference between these two areas, that is the area shaded, 
represents therefore the work done by the steam or gas upon the 
piston in one double-stroke of the latter, i.e. in one revolution of 
the engine shaft. Therefore the mean height of this diagram, 

i.e. p^ , represents the mean effort. The indicator is 

calibrated so that by multiplying the mean height of the diagram 
in inches by a constant we get at once the mean pressure acting 
on the piston in lbs. per sq. in. Let this mean pressure be ^^ lbs. 
per sq. in. 

Now let A be the area of the piston in square inches ; L the 
stroke in feet and N the number of revolutions per minute of 
the engine shaft. N is measured during the test by a counter. 

Then E^ = p^A, 

:. Work done per revolution = E^ . L = Pm^L, 
.*. Work done per minute = pJf^ALN. 

:. Indicated Horse-Power = qq aaa (^)' 

Brake Horse-Power. The Brake Horse-Power of an engine is 
the power output or as it is sometimes called the Effective Horse- 
Power. It is given its name because it is usually measured in 
tests by an arrangement called a "brake,'* a simple form of which 
is as follows. A rope J5, Fig. 46, is passed over the flywheel A ; it 
is usually made up of three or four pieces of rope knotted together 
at the ends and held apart by distance pieces D. On the side on 
which the flywheel would tend to lift it is hung a weight pan C 
which is often provided at the bottom with a piece of rope secured 
to the floor to prevent the weights from being bodily carried 
right over the flywheel. The rope is connected at the other end 
to a spring balance the reading of which may vary slightly from 
time to time. 

As the flywheel rotates in the direction of the arrow, the rope 
will slip continuously. We have here exactly the converse of 
the belt drive of a pulley. Here the pulley is the driving member, 



IV] 



MACHINES AND EFFICIENCY 



«79 



and it spends its energy in overcoming the friction or grip betv^een 
the rope and the flywheel. Now let the weight on the pan be 
W lbs. and the reading of the spring balance t(; lbs , and let r feet 
be the radius of the flywheel. 



■d=2r 




w/y/^y 



Fig. 46. 



In one revolution the flywheel does an amount of work 
equal to 

E =- 27rrW - 2iTrw = 27rr (W - w) = ird {W - w). 

If therefore the flywheel makes N revolutions per minute we 
have 

Work per minute output of engine = EN = irdN {W — w); 

EN 



:. Brake Horse-Power 



I.e. B.H.p. = 



33,000' 
7rdN{W-w) 



(2). 



33,000 

The diameter d should be measured to the centre of the rope. 
The ratio ' is called the mecJianical efficiency of the engine. 

Numerical Example. In the test of a steam-engine the 
mean pressure was found from the indicator diagram to be 60*3 lbs. 
per sq. in. and the stroke was 12 inches. The piston was of 10 inches 
diameter and the number of revolutions per minute uxis 122. The 



\/ 



80 MACHINES AND EFFICIENCY [ch. 

diameter of the flywheel was 6 feet and the rope was 1 inch in diameter, 
and the weight W was 240 lbs., the spring balance reading being 4 lbs. 
Find the i.h.p., b.h.p. and mechanical efficiency. 

Area of piston = A = ^x 10^= 7854 sq. ins., 

p^=60-3, L=ltt.; 

_ 60-3 X 78-54 x 1 x 122 

. . I.H.P. - gg^^^^ 

= 17-5, 

^•^•^'= 33,000 ' ^=5+j2=508ft. 

ff X 508 X 122 X 236 
33,000 
= 13-9, 

Mechanical efficiency = -^—Ll = — -— = '794 
'^ I.H.P. 17*5 

« 79*4 per cent. 



SUMMARY OF CHAPTER IV. 

A machiiie is an appliance for receiving energy from some out- 
side source and converting it into some more convenient form, 

-,-.,, ^ Resistance 

Mechamcal advantage = — :p^ — - — . 

. . . _ Distance moved at the effort in a given time 

Distance moved at the resistance in a given time * 

The efficiency of a machine is the ratio of the energy that it gives 
out to the energy that it receives. 

Vffi ■ - Ideal effort 
~" Actual effort ' 

Mechanical advantage = velocity ratio x efficiency. 

If a machine is " self-sustaining '* or not reversible, its efficiency 
cannot be as much as 50 per cent. 

Work done upon rotating bodies per revolution in ft.-lbs. 
= 27r X torque in lb. -ft. 



ivj MACHINES AND EFFICIENCY 81 

Indicated Horse-Power (i.h.p.) of an engine = ^ ^^^ , 

oo,\J\)\) 

_ , ,T T^ / X £ • 7rdN{W-w) 

Brake Horse-Power (b.h.p.) of an engine =^ 



Mechanical efficiency 



33,000 

B.H.P. 



I.H.P. 



EXERCISES. IV. 

1. In a wheel and axle the diameter of the wheel is 3 ft. 6 ins. 
and the diameter of the axle is 10 ins. The diameter of the rope 
attached is in each case 1 in. Find the weight which can be lifted 
by a pull of 50 lbs. on the rope attached to the wheel. 

2. If in the last example a weight of 195 lbs. is lifted what is 
the efficiency of the machine? 

3. The diameter of the wheel in a wheel and axle is 18 ins., 
and that of the axle 5 ins. Neglecting friction what pull on the 
wheel will raise a weight of 600 lbs. ? If it requires a pull of 
200 lbs. weight to lift this load what is the efficiency? Also find 
the mechanical advantage and velocity ratio of the machine. 

4. Find the h.p. of an engine which will raise 1000 gallons 
of water per min. from a depth of 240 ft. The efficiency of the 
engine is 55 per cent. 

5. The inclination of a plane is 3 in 5. Find what force acting 
parallel to the plane will support a load of 2 tons neglecting friction. 
Also find the force which would be required acting parallel to the 
base of the plane. 

6. The handle of a lifting jack measures 24 ins. in length and 
the pitch of the screw is f in. What force applied at the end of 
the handle would be required to raise a load of 22 cwts., the effect 
of friction being neglected? 

7. A shaft transmits 50 h.p. at 250 revs, per min. Find the 
twisting moment in inch-lbs. 

8. The twisting moment on an engine shaft is 20,000 in. -lbs. 
and it makes 180 revolutions per min. Find the h.p. transmitted. 

9. The pitch on a screw-jack is ^ inch, the distance from the 
axis of the screw to the end of the handle 26 inches. Find the 
velocity ratio. If the law is F = -03 TF + 9*45, find the load which 
will be lifted by a force of 56 lbs. wt. applied at the end of the handle. 
Find also the efficiency at this load. 

A. M. 6 



82 



MACHINES AND EFFICIENCY 



[CH. IV 



10. The diameter of a steam-engine cylinder is 9 ins., the length 
of crank 9 ins., the number of revolutions per min. 110, and mean 
effective pressure of the steam 35 lbs. per sq. in. ; find the indicated 

H.F. 

11. In measuring the brake h.p. of an engine a rope passes 
round the flywheel, one end being fixed to a spring balance; the 
other end carries a weight of 120 lbs. If the wheels make 150 
revs, per min. and the spring balance indicates 15 lbs. what is the 
H.p. transmitted ? The fljrwheel is 5 ft. in diameter. 

12. A steam pump is to deliver 1000 gallons of water per 
minute against a pressure of 100 lbs. per sq. in. Taking the efficiency 
of the p\unp to be '10, what indicated h.p. must be provided? 

13. The diameter of the cylinder of a double acting engine 
is 10', stroke 15'% number of revolutions per min. 120, and the mean 
steam pressure 48 lbs. per sq. in. Find the h.p. transmitted. 

14. In a rope-brake dynamometer the diameter of the brake 
wheel is 10 ft., rope is 1|* diameter, weight on rope at one end 
is 200 lbs. and pull on spring balance at the other end is 18 lbs. 
weight. If the wheel makes 90 revs, per min. find the h.p. trans- 
mitted. 

15. The following results were obtained in a test of a steam- 
engine: 1 H.p. = 7'7; revs, per min. = 164; diameter of brake- 
wheel 3 ft. ; diameter of rope ^ in. ; weight on brake 150 lbs. ; 
reading of spring balance 2'6 lbs. Find the mechanical efficiency 
of the engine. 

16. The following results were obtained in a test of a machine 
whose velocity ratio = 8: 



Load or reEdstance lbs. 





6 


10 


20 


30 


40 

1 


Effort lbs. 


30 


4-8 


60 


100 


140 


18-0 



Plot a curve showing the efficiencies at various loads and find 
the efficiency for a load of 25 lbs. 



CHAPTER V 

VELOCITY AND ACCELERATION 

What do we mean when we say that a train is going at 
60 miles an hour at a certain point ? We do not mean that in one 
hour the train actually goes 60 miles; it might stop altogether 
after it has gone 20 miles. But what we mean is that if the train 
continued to move at the same speed or velocity for one hour 
it would then have gone 60 miles. Expressing velocity in scientific 
language we say that "velocity or speed is the rate of change of 
position or space with respect to time." 

Velocity is a vector quantity * ; its direction is of importance 
as well as its magnitude. It is here that many people use the 
term "speed" and "velocity" with a slight difference of meaning. 
When speed is spoken of the direction does not come into con- 
sideration but velocity involves the direction of the motion. 

Velocity of a point. When a body is moving, different points 
in it may be moving with different velocities, so that in strict 
language we do not speak of the velocity of a body but of the 
velocity of a point. 

Uniform Velocity. The velocity of a point is said to be 
uniform when it maintains the same direction and magnitude 
(i.e. the point passes through equal distances in the same direction 
in equal times). 

Suppose that a point has a uniform velocity of 10 feet per 
second in a certain direction. Then in 1 second it will move 
through 10 feet ; in 100 seconds it will move through 1000 feet ; 
in yJ^ second it will move through ^V^^ ^^ ^ ^^^* ^^^ ^^ o^- 
This is expressed in symbols as follows : If a point has a uniform 

• Cf. p. 1. 

6-2 



84 VELOCITY AND ACCELERATION [ch. 

velocity v feet per second, then the distance s in feet covered in 
t seconds is given by the formula 

8 = Vt (1). 

This is true no matter how large or small t may be. 

Variable Velocity. In practice velocity is seldom if ever 
uniform although it may over a certain time be sufficiently 
nearly so to be reckoned as uniform for all practical purposes. 

There are two causes that may disturb uniformity of velocity : 

(a) Variation in magnitude. 

(b) Variation in direction. 

Very often these two variations occur together, but for the 
rest of this chapter we will consider change in magnitude only. 

Velocity variable in magnitude. Suppose that the times 
are recorded at which a moving point passes certain stations 
and that the distances of these stations from a suitable starting 
point are plotted against the times of passing. The curve CPQD, 
Fig. 47, obtained by joining up the points is called the space 
curve. 

At the instant from which the time is reckoned the distance 
from the starting point is AC; then at any point such as P, after 
a time t = AT has elapsed, the moving point is at a distance 
s = PT from its starting point. 

Now consider a point Q on the space curve very near to P, 
and let PR be drawn perpendicular to QU; while the point has 
moved a distance equal to QU — PT = QRj the time has increased 
by an amount TU = PR. 

T^T QR Distance moved . ^ 

Now _^=____g__=tane. 

Next suppose that the points Q and P move closer and closer 
to each other ; the line PQ then gradually approaches the position 
of the tangent XY shown in dotted lines, and the slope of this 
tangent may be taken as tan 6 if PQ is sufficiently small. 

Now we define the velocity at any point as the value which 

^, distance moved , .i. j- . j i 

the — -' T-i approaches as the distance moved becomes 

time taken 

smaller and smaller. It follows from this that the slope o/ the 



V] 



VELOCITY AND ACCELERATION 



86 



tangent to the space curve at any point measures the velocity at that 
point. 

In working from the diagram we must be careful to allow 
properly for the scales ; referring to the figure we have 

,^ , ., , , ... /I YZ on space scale 

Velocity at given point = tan 6 = ^ ^ ^n time scale " 



« 
O 

o 

^ 




Time (t) 



Pig. 47. Space Curve. 



Numerical Example. The folloming results were obtained in 
timing a man walking over a certain distance. Find the velocity 
at the commencement and after 40 minutes from the start of the tesf. 
Find also the average velocity over the whole test. 



Time in seconds 





10 


20 


30 


40 


50 


60 


Distance in feet from 
starting point 


160 


196 


263 


345 


440 


505 


550 



The results are plotted in Fig. 48 where BCD is the apace 
curve and we are required to find the velocities at B and G. 
To do this we have, as previously explained, to draw tangents 
to the space curve at B and C. This is not easy to do accurately ; 
so far no graphical construction is known which is mors accurate 
than drawing a line by eye to touch the curve. 



86 



VELOCITY AND ACCELERATION 



[CH. 



rvu ^7 1 -^ ^ » + /3 ^^ 241 feet 
Then Velocity at 5 = tan ^, - -^^ » 60 seconds 



401 feet per second. 



o 
o 












/ 
, / 

/ 

/ 


o 












'/ 


s 










/ 












f 


r ' ' 

J 




o 








/J 






"* 




■ 


/ 


'L^f 


ace 


y 
y 

y 


o 






y / 
// 
// 


Our 

y 


/C y 

y 


V 


CO 




/ 


• 








o 




'7 / 


y 








B 

H 

o 


yk ^ 


[^. 










/^ 


J 








. _ — ' — - 











































fC 






10 20 30 40 50 60 

Seconds 

Fig. 48. 

It is common to measure some velocities in miles per hour. 
Now 1 mile per hour = 5280 feet in 3600 seconds 

5280 22 , ^^_ , ^ , 

•" o/3?m = TE " 1"467 feet pel' second r 



V] VELOCITY AND ACCELERATION 87 

401 
/. Velocity at 5 in miles per hour— , ,^„ = 2*73 miles per hour, 

Velocity at (7 = tan 0^ = j^^ = -^ = 8 feet per second 

g 
«= ■■ .^- = 5*45 miles per hour. 

Average velocity = -^^ = -^ = 6*67 feet per second 

■= ^ = 4-65 miles per hour. 

A useful figure to remember is that 60 miles an hour is equal 
to 88 feet per second, or one mile per hour = f| feet per second. 

Velocity Curve and its relation to the Space Curve. Next 
suppose that we know the velocity at every time and that we 
plot velocities upon a time base; then the resulting curve is 
called a velocity curve QHJK, Fig. 49. AO represents the 
velocity at the beginning of the period of time under consideration 
and is called the initial velocity and will be given the letter u. 

Now consider the relation between this curve and the space 
curve plotted on the same base (to save confusion it is preferable 
to plot one diagram above or below the other). We have already 
shown that the velocity v at the middle of a very short time LM 
is equal to the slope of the tangent to the space curve at the 
corresponding point. Since LM is so very short, the slope of 
this tangent is given by 

QB QR 



tan^ = 



PR LM' 
QR, 



"''-LM' 

.*. QR = LM X V = Area of shaded strip of the velocity 
curve. But QR is the increase in the ordinate of the space curve, 
and we could show similarly for any other strip that the increase 
in the ordinate of the space curve represents the area of the 
corresponding strip of the velocity curve. 

.*. QU = Total increase in space from the beginning = Area 
of velocity curve from A to M, i.e. area AGJM. But this is 
exactly the relation which we have explained (p. 43) betwee^i 
^ slope curve and its primitive curve. 



98 VELOCITY AND ACCELERATION [oh 

Therefore the space curve is the sum curve of the velocity curve. 




Time 



Fig. 49. Velocity and Space Curves. 

We may use as a general rule the following relation which we 
have proved above. If a curve A is the sum curve of a curve 
B, the ordinate of B at any point represents the slope of A at the 
corresponding point. 

Some special cases of Velocity and Space Curves, (a) Constant 
velocity. If the velocity of a point is constant, the velocity curve 
is a horizontal straight line, Fig. 50. If the sum curve construc- 
tion be carried out for this we get a sloping straight line AD^ 
assuming that we commence reckoning our distances from the 
point at which the time commences. This is because all the 
mid-ordinates of the velocity curve when projected horizontally 



V] 



VELOCITY AND ACCELERATION 



89 



come to the point G so that all the elemental pieces of the sum 
curve are parallel to PG. [If, instead of taking our distances from 
the point at which the time commences, we take them from some 
other point, we shall get a space curve such as cR parallel to 
ADy but in all further cases we shall assume that the space is 
considered as commencing at the beginning of the time interval.] 




Fig. 60. Constant Velocity. 

We then have : distance covered from AtoB 
= area AGKB = vt, 
i.e. s — vt 



(1). 



Since for any value of t the space s is equal to vt, the space 

curve is a straight line such that tan DAB = - = v. 

I 

(6) Velocity increasing uniformly. If the velocity increases 

by the same amount in each unit of time we shall obtain for our 

velocity curve a sloping straight line GKy Fig. 51. 




Fig. 61. Velocity increasing uniformly. 

If we apply to this the sum curve construction the space 
curve will be found to be a parabola ARD. 



90 



VELOCITY AND ACCELERATION 



|_CH. 



We then have : distance covered from ^ to jB 
- area AGKB = i AB {AO + BK) 
= lt{u + v), 



I.e. 



8=t 



m 



(2). 



We shall see later that it is sometimes more convenient to 
write this , 



^ (v—u)t 



(3). 



(c) Velocity decreasing uniformly. In this case the velocity 
curve GK, Fig. 52, will also be a straight line but will slope 




Fig. 52. Velocity decreasing uniformly. 

downwards. The space curve will also be a parabola ARD, but 
it will curve the opposite way from the previous case. 
After a time t therefore we get 
8 = area AGKB 

= ^ (w + v) as before 
'^ ^{u+u- (u-v)} 

^^^M-^^-^t (4). 



V] 



VELOCITY AND ACCELERATION 



91 



Numerical Example. A point starts from rest and increases 
its velocity uniformly for 10 seconds at the end of which it has 
a velocity of 10 feet per second. It continues to move for 10 
more seconds at this velocity and the velocity then diminishes 
uniformly for 5 seconds when it comes to rest. How far has it 
travelled ? 

The velocity curve for this case is as shown in Fig. 53. For 




20 B 
Pig. 53. 

the first 10 seconds between A and (7 it is a sloping straight 
line AG\ for the next 10 seconds the velocity is constant so 
that the velocity curve is a horizontal straight line CD and for 
the next 5 seconds the velocity falls uniformly to zero so that 
the velocity diagram is the sloping straight line DB. 

Now the total space covered in the 25 seconds vdll be repre- 
sented by the area AGDB. 

ThLs can be estimated as follows : 

Area of A ACH = l^^i? = 50 feet. 



Area of rectangle GDJH= 10 x 10 

10x5 



Area of A DJB = 



2 



100 feet, 
25 feet. 



Woi^\ distance covered =175 feetc 



92 VELOCITY AND ACCELERATION [ch. 

The space curve will be as indicated, AE and FQ being 
parabolic arcs and EF a straight line. As an exercise the reader 
should draw the curve by the sum curve construction. 

Suitable scales would be as follows : time 1" = 5 seconds ; 
velocity 1" = 5 feet per second ; polar distance = 2 inches. 

Then the space scale will be 1'' = 2 x 5 x 5 = 50 feet so that 
BG should measure 3'5 inches. 

Acceleration. When the velocity of a body is changing it is 
said to have an acceleration. Acceleration is measured by the 
rate of change of velocity and we have already explained that 
change of velocity may take place in magnitude or direction or 
both. For the present we will confine ourselves to change of 
magnitude of velocity and will assume that the direction remains 
constant. 

Suppose that a body is moving at a certain instant with a 
velocity of 10 feet per second and that one second later it is 
moving in the same direction with a velocity of 12 feet per second. 
In one second the velocity has gained by 2 feet per second, so 
we say that the mean acceleration is 2 feet per second per 
second; this is often written for brevity 2ft./sec.2 

Now let the velocity curve be LMNK, Fig. 54. At the time 
represented by the point T the velocity is represented by TM 
and after a short time T?7 it is represented by UNy so that in 
time TU the point has gained in velocity by an amount NO. 

NO 
:. Mean velocity gained in unit time = ^ttj- 

Now the points TU are very close together and as N comes 
closer still to M the line joining MN ultimately becomes the 
tangent XX to the velocity curve at l-he point M, 

Then rate of change of velocity 

NO 
«= Y77) = slope of tangent XX = tan 9. 

Therefore the acceleration at any point is represented by the 
.'lope of the velocity curve at the given point. 

If we obtain the accelerations at a number of points and plot 
them against the times, the resulting curve will be an acceleration 
curve. 

Positive and negative acceleration. We have up to the present 



V] 



VELOCITY AND ACCELERATION 



93 



only spoken of velocity gained but the term "gain" must be 
considered as including "loss" and when there is a loss we shall 
regard it as a negative gain. Returning to our numerical 
illustration suppose that instead of being 12 feet per second at 
the end of one second the velocity is 8 feet per second ; in one 
second the velocity has lost 2 feet per second and we should say 
that the mean acceleration 18—2 feet per second per second. 




Fig. 54. Acceleration. 

Now when the velocity is decreasing the tangent, such as YT, 
Fig. 54, cuts the base at a point in advance of the point of contact 
whereas when the velocity is increasing the tangent cuts the 
base behind the point of contact. This enables us to formulate 
the following rule. If the tangent to the velocity curve cuts the 
time base at a point behind the point of contact, the acceleration 
is positive and if it cuts at a point beyond the point of contact 
the acceleration is negative. 

General relation between Acceleration, Velocity and Space 
Curves. We have shown that the slope at any point of the 
velocity curve determines the acceleration and we have previously 
shown that the slope at any point of the space curve gives the 
velocity; there is therefore the same relation between the 
acceleration and velocity curves as there is between the velocity 
and the space curves. We get therefore the following very 
important rule. 

The velocity curve is the sum curve of the acceleration curve and 
the space curve is the smn curve of the velocity curve. 



94 



VELOCITY AND ACCELERATION 



[CH. 



This is illustrated in Fig. 65 in which, to save confusion, the 
three curves have been drawn upon separate bases. CDE is the 
acceleration curve ; drawing a sum curve with polar distance p^ , 
we get the velocity curve AFO if the point starts from rest; if 
the point has an initial velocity u we set up A A' = uon the velocity 
scale, obtained as described later, and start the sum curve at 




Fig. 55. Relation of Acceleration, Velocity and Space Curves. 

A' thus obtaining the velocity curve A'F'O' shown in dotted 
lines. Drawing the sum curve of this with a polar distance p^ we 
get the space curve AH J. 

Scales. Suppose that the time scale is I'' = a; seconds, and 
that the acceleration scale is V = y ft./sec.^ and suppose that 
Pi is measured in actual inches. Then the velocity scale will be 
V = PiXy ft./sec. Now let p^ be also measured in actual inches. 
Then the space scale will be 

V = PzVi^^y ^®®** 

This may be explained as follows. 

One square inch of the acceleration curve represents xy units 



V] 



VELOCITY AND ACCELERATION 



96 



of velocity and the sum curve construction gives the area divided 
by the polar distance so that one inch on the sum curve AFG 
represents fyxy = z say. 

By similar reasoning one inch on the sum curve AH J represents 
P2XZ = PiPzX^. 

As a numerical illustration let the time scale be I'' = 10 
seconds and the acceleration scale V = 2 feet per second per 
second and let Pi = 2 inches ; then the velocity scale will be 

1" = 2 X 2 X 10 = 40 feet per second. 

Next let P2= li inches ; then the space scale will be 

1^ = IJ X 40 X 10 = 600 feet. 

By a careful choice of the polar distances pj^, p2 we can 
obtain convenient scales; we should, for instance, have done 
better in the above case to have taken pi = 2J inches and 
2)2 = 2 inches, our velocity scale would then be I*' = 60 feet 
per second and the space scale V = 1000 feet. 

Constant A cceleration ; equations of motion. If the accelera- 
tion is constant, the acceleration curve is a horizontal straight 




Fig. 66. Constant Acceleration. 

line CD, Fig. 56 ; the sum curve, i.e. the velocity curve of this, 
will be the sloping straight line HE^ while the space curve AFO 
will be a parabola, the sum curve of a sloping straight Une being 
a parabola. 



/ 



96 VELOCITY AND ACCELERATION [ch. 

From these curves we can deduce the following formulae: 
KE = area of acceleration curve = ^t ; 
:.v=BK-\-KE 

= u-\- at (5). 

s = area AHEB 

{u+ v) /u -h u + at'' 



= t 



_ /u-\- u+ at\ 

" ' V 2 ; 



2 
= ut+lat^ (6). 

We can get a third relation as follows : 
By squaring equation (5) we have 

^;2 = {y2 + atf = u'^-\- 2uat + aH^ 

= u^ + 2a {ut + lat^) 

= u^-{-2as [frpm (6)] (7). 

These equations (5) to (7) are often called the equations of 
motion and are very useful in problems in which the acceleration 
is constant. 

Numerical Examples. (1) A point moves along a straight 
line under an acceleration of 10 ft.jsec.^ The initial velocity ig 
1 Jt.jsec, What is the velocity after it has passed over 12 feet? 

In this case w = 7 feet per second, 

a = 10 feet per second per second, 
s = 12 feet. 

Therefore using equation (7) 

^2 = 72 + 2 X 10 X 12 

= 49 + 240 

= 289, 
V = \/289 = 17 feet per second. 

\£) A train is running at 20 miles an hour and is stopped by 
brakes in 10 seconds, the retardation being constant. At how many 
yards from the stopping point were the brakes applied ? 

60 miles an hour = 88 feet per second. 

88 
/. 20 miles an hour = -^ feet per second. 

«5 



vj VELOCITY AND ACCELERATION 97 

88 
In this case u= -^ft./sec, t= 10 and v=0; 

.\0= ~+lOa [from (5)]; 

.*. a=-g^ft./sec.2; 

880,, ,, 440, ^ 440 , 
= -3- (1 - J) = -3- feet = -^ yards 

= 48-89 yards. 

Gravity Acceleration "g." If bodies are allowed to drop 
freely they will be found to have an acceleration which is 
practically constant. 

This acceleration is called the gravity acceleration and is 

given the letter g. Its value varies slightly with the latitude and 

with the height above sea-level and in London is usually taken 

as 32*2 feet per second per second. We will indicate later an 

interesting simple experiment for determining g and will now 

derive simplified formulae for the case of bodies falling freely 

from rest. In equations (5) to (7) therefore we have w = and 

a = g and it is usual to replace the distance or space s by the 

height h. Our formulae therefore become 

)/ 
'^-gt (8), 

h = igt^ (9), 

v^ = 2gh (10). 

Formula (10) is of the greatest possible importance and may 
be rewritten in the forms 

v=^sf2gh (11), 

'^=r, (12). 

The student must make himself absolutely familiar with these 
formulae and should not feel fully satisfied until he can work 
successfully through all the exercises at the end of the present 
chapter. 

A.M. 7 



.•.2-9=y, 



98 VELOCITY AND ACCELERATION [ch. 

Numerical Example. A stone is let fall down a well and the 
splash is heard 2-9 seconds later. If the time for the sound to travel 
to the top of the well be neglected, what is the depth of the well ? 

Let h be the depth of the well. 

Then the time in dropping is obtained by equation (9) 

h=yt^, i.e. t= ^ —; 

but t = 2-9 seconds, 

2h 

32-2* 
Therefore, squaring, 

2-92 = Jt_ 
^^ 161' 

h= 16-1 X 2-9^ = 135 feet approx. 

With what velocity must a stone be projected if it is to reach a 
height of 120 feet ? 

Here we have in our general equation 

y^ = u^ -{. 2as, V = Of a = — g, s = h, 
/. = w2 _ 2gh, 
u= sl2gh 
= n/2 X 32-2 X 120 = 88ft./3ec. approx. 

Limits of use of simple formulae. In using these simple 
formulae care must be taken to remember that they are based 
upon the assumption that g is constant and that the bodies fall 
"freely," i.e. that the air resistance is negligible. 

As a matter of fact the air resistance is appreciable for great 
heights with light bodies ; but for this fact a rain drop in falling 
from a cloud would acquire such a high velocity that it would 
kill a man if unprotected by armour. Moreover, if the height is 
very great the body will not fall vertically, judged by standards 
upon the earth. This point was illustrated in an interesting 
manner in some experiments which were carried out in a deep 
vertical mine shaft in the United States of America, one of the 
shafts being 5300 feet deep. 

Smooth metal balls 2 inches in diameter were suspended by 
threads and allowed to drop by burning the thread, a box of 
clay being placed 4200 feet beneath. All the balls struck the 
east wall of the shaft before reaching the box. This was due to 



V] VELOCITY AND ACCELERATION 99 

the movement of the earth from west to east, this movement 
being sufficient to cause the balls to be struck by the east wall 
before they came to the box. In one case 800 feet of fall was 
sufficient to make a ball dropped 4 feet from the east wall strike 
against it. 

Distance moved in a particular second. In several problems 
upon velocities and accelerations we require to consider the 
distance moved in a particular second under a constant velocity. 

Suppose for instance that we want to know the distance 
moved through in the fifth second. In five seconds it will have 
moved through a certain distance Sg, given by, putting t = 5 in 

equation (6), 25a 

8,= 5u+-^ (13). 

In four seconds it will have moved through a distance s^ given 

54=4w+-2~ (14). 

Now the difference between the distances moved in five and 
four seconds respectively must give the distance in the fifth 
second, so that we have 

Distance moved in fifth second = 55 — 54 

, ^^ /ir\ 

= '^+2" ^^^^• 

Now take the most general case. It is clear from the above 
illustration, which could be employed for any numerical value, 
that the distance moved through in the nth. second must be the 
difference between the distances moved through in n and (w — 1) 
seconds respectively, 

i.e. Distance moved through in Tith second 

= 5„ — S„-i 

= {un + Jari^} — {u{n— l)-\- \a {n — 1)^} 
= {un + Jan^} — {un — u -\- \a {n^ — 2n -\- 1)} 

= un-{- \an^ — un-\-u— Ian- + —^ — ^ 
2an a 

= ^ + 1(2^-1) 

= w + a (w - i) (16). 

7—2 



100 VELOCITY AND ACCELERATION [ch. 

Now {n — J) is the time to the middle of the nth second. 
Therefore by equation (5) the velocity at that instant 

= V = u ■}- at 

= u -\- a {n — J). 

We thus obtain the very useful rule that : The space in feet 
moved through in any particular second is the velocity in feet per 
second at the middle of that second. 

We have proved this result in the above manner to give us 
an exercise in reasoning by the manipulation of formulae but we 
could have proved it, perhaps more simply, from a consideration 
of the velocity diagram as follows : referring to Fig. 51 let NQ 
represent the ordinate of the velocity diagram at the end of 
n seconds and let LM represent it at the end of (n — 1) seconds, 
so that MQ represents the nth second. Since the space curve is 
the sum curve of the velocity curve the increase TU in the space 
over this second is represented by the area of the strip LNQM 
of the velocity curve and this is equal to 1 x mid -ordinate = 
velocity at the middle of the second under consideration. 

Numerical Example. A train in two successive seconds moves 
through 20-5 and 23-5 feet respectively. If it is being accelerated 
uniformly what is its acceleration and what was the velocity at the 
beginning of the first second ? 

Suppose that u is the initial velocity. 

At the end of the first second we have 

s = ut -i- lat^, 

20-5 = w.l + Ja . 1 (1). 

At the end of the second second we have 

(20-5 + 23-5) = ?t . 2 + Ja . 22 (2), 

.-. 20-5= u + ^ from (1), 

235 = ?t 4- -^ by subtracting (1) from (2) ; 

.*. 3 = a by subtraction ; 

.-. u = 20 5 - ^ = 20-5 -1-5=19; 

i.e. Acceleration ^ 3 ft. /sec. 2, 

Initial velocity = 19 f t./sec. 



V] VELOCITY AND ACCELERATION 101 



SUMMARY OF CHAPTER V. 

Velocity is the rate of change of position with respect to time. 

Acceleration is the rate of change of velocity. 

The velocity of a point is measured by the slope of the tangent 
of the space curve. 

The acceleration of a point is measured by the slope of tho 
tangent of the velocity curve. 

For bodies moving under constant acceleration, 

V = u + at, 

s = ut + \at\ 

For bodies starting from rest imder gravitational acceleration, 

^=^ 

g 

In all problems it is better to reason out as much as possible 
from first principles than to attempt to remember the formulae 
and to apply them directly. 



EXERCISES. V. 

1. What will be the velocity of a body after falling 25 ft. from 
rest? 

2. Find the average speed of a train wliich runs from London 
to Grantham, a distance of 105^^ miles, in 1 hour 55 minutes. 

3. A stone takes 2% sees, to drop to the bottom of a well. What 
is the depth of the well? 

4. Suppose a body to have fallen h feet in t sees, from rest 
according to the law h = 16-li'. Find how far it falls between 
the times t = 3 and t = 3-1 ; between t = 3 and t = 3-01 ; between 
^ = 3 and t = 3-OOL Find the average velocity in each of these 
intervals of time. What do we mean by the actual velocity when 
Ms 3 sees. ? 

5. What is an acceleration of 60 miles per hour per minute 
in feet per sec.^? 



102 



VELOCITY AND ACCELERATION 



[CH. V 



6. X and t are the distance in miles and -the time in hours of a 
train from a railway terminus: 



X 

t 




' 

i 


1-5 


60 


140 


190 


21-0 


21-5 


2r8 


23 


24-7 


26-8 


01 


0-2 


0-3 


0-4 


0-5 


0-6 


07 


0-8 


0-9 


10 



Plot on squared paper. Describe why it is that the slope of 
the curve shows the speed. What is the greatest speed in this 
case and where approximately does it occur? 

7. The following numbers give v the speed of a train in miles 
per hour at the time t hrs. since leaving a railway station. Draw 
a diagram showing the distance covered at the various times and 
find the total distance covered. 



V 





2-4 


4.7 


7-2 


96 


120 


14-3 


t 


•00 


004 


008 


•12 


•16 


•20 


24-9 


V 


16^9 


18-9 


20-7 


22-2 


23-4 


24-3 


240 


i 


•28 


•32 


•36 


•40 


•44 


•48 


•52 



8. Express 2 ft. per sec. in cms. per min. 

9. A train which has constant acceleration starts from rest, 
and at the end of 3 sees, has a velocity with which it would travel 
through 1 mile in five mins. Find the acceleration. 

10. A train goes from one station to another 5 miles off in 
8 mins., first moving with constant acceleration and then with 
an equal retardation. Find its greatest speed. 

11. A train reduces speed from 45 miles an hr. to 15 miles 
an hr. in 800 yds. How much farther will it go without stopping ? 

12. A point starting from rest passes over 121 ft. in the sixth 
second. What is the acceleration? 

13. From a balloon which is ascending with a velocity of 32 ft. 
per BBC, a stone is let fall and reaches the ground in 17 sees. 
How high was the balloon when the stone was dropped ? 

14. A train goes a distance of 120 miles in 3 hours. During 
the first hour the speed rises uniformly from rest ; during the second 
hour it remains constant ; and during the third hour it falls uniformly 
to rest. What is the speed during the second houi? 



CHAPTER VI 

VELOCITY CHANGE IN DIRECTION ; RELATIVE VELOCITY 

We have considered so far only the case of motion in a straight 
line and have taken into consideration only changes in magnitude 
of the velocity; but we may also have change in direction, 
with or without change in magnitude as well. The case of a body 
moving with constant velocity in a circle is an example in which 
the magnitude of the velocity is constant but the direction is 
constantly changing. 

Combination of Velocities. The actual velocity possessed by 
a point may be the combination of two or more velocities, and as 
velocities are vector quantities, they are added together in 





Fig. 57. Combination of Velocities, 

exactly the same way as forces, i.e. by the law of vector addition. 
Suppose for instance that we are standing at one end A (Fig. 57) 
of a railway carriage moving with a velocity v^ and that we walk 
across the carriage with a velocity Vg ; then our actual velocity 
will be the combination of the velocity v^ of the train itself and 
of our own velocity v^, i.e. Vj. in the direction AG by the law of 
vector addition. 

A good familiar example in which a body has a velocity 
compounded of two velocities is to be obtained from the case of 
a wheel rolling along the ground. Any point on the wheel is 
moving around the axle and the axle is at the same time moving 
along parallel to the ground so that each point upon the wheel is 
actually describing a curved path — indicated in dotted lines in 



104 



VELOCITY CHANGE IN DIRECTION; 



[CH. 



Fig. 58. This curved path is called the cycloid and is also used 
by engineers in considering gear teeth. This curve can be drawn 
b}^ rolling a half-crown along a ruler and resting a pencil against 




Fig. 68. Wheel rolling along the ground. 

the edge of the coin. A milled coin such as the half-crown is 
better than a penny because it will not slip on the ruler. With 
a little practice a very smooth curve can be obtained. 

Change of Velocity. Suppose that a point at A, Fig. 59, 
has a velocity v^ at one instant and after a certain time it is at 





y 



A' a i)^ 

Fig. 59. Change of Velocity. 

B and has a velocity Vg. Then the change of velocity v^ is 
defined as the velocity which would have to be compounded or 
combined with Vj to give v^. That is Vg is the resultant of v-^ 
and Vq , or expressing this in vector notation we have 

^2 = 2^1 + ^0- 

This problem arises in engineering calculations in considering 
the impact of water upon the vanes of a water wheel or turbine. 

Numerical Example. A jet of water moving with a velocity 
o/ 80 feet per second impinges upon a curved plate and has its 
direction turned through 120°, without altering its magnitude. 
What is the change in velocity ? 



VI] RELATIVE VELOCITY 105 

Referring to Fig. 60, we draw a6 to a suitable scale to represent 
80 feet per second and ac at 120° to it to represent 80 feet per 
second also and then join hc\ then he represents the change of 
velocity and if the diagram is drawn to scale be will be found by 





80 



measurement to give about 138-6 feet per second. To find v^ by 

calculation without actually drawing the triangle to scale we 

draw ad perpendicular to he, we then note that 

cd = ac cos 30° = 80 x -866, 

.*. he = 2cd = 160 X '866 = 138-6 feet per second nearly . 

Relative Velocity. We now come to a very important 
portion of the subject which students often find rather difficult 
to understand and to which therefore we wish to give particular 
attention. In the ordinary way when we speak of velocities in 
a certain direction (say 4 miles an hour in a northerly direction) 
we leave out of consideration the fact that the earth is not 
fixed. Since the earth itself is rotating on its axis as well as 
moving through space at a very high velocity the actual velocity 
of any point is the combination of the velocity commonly referred 
to and that of the earth. We express this by saying that 
velocities as ordinarily measured are relative to the earth. 

If we sit at the back of a dog-cart we can easily get the idea 
that the road is moving away from under us; that is because 
we regard ourselves as fixed and therefore relatively the road is 
moving away from us. 

If, again, two trains are standing alongside in a railway 
station, and one starts moving, a person sitting in one train and 
looking at the other always has some doubts as to which of the 
trains is moving. Suppose that we are sitting in the train which 
we wiU call A and that the other is B. If B moves we have the 
sensation oi moving in the opposite direction. 



106 



VELOCITY CHANGE IN DIRECTION; 



[CH. 



Now if two bodies A and B are both moving, the velocity of 
B relative to A is the velocity which B would appear to have if A 
were regarded as stationary. 

Let us consider a fact that every observant reader will 
already have noticed, viz. that the rain splashes upon the 
window of a train or other moving vehicle are never vertical 
although the rain may be falling quite vertically; they are 
always inclined away from the direction of the train, i.e. they 
always seem to be coming towards the train as indicated in 
Fig. 61. In answer to the question as to why this is, we usually 
say that the relative velocity of the rain to the train is in that 
inclined direction; but that answer does not give very much 




Apparent 



/elocity 
of Rain 



Actual 
X Velocity 
* of Rain 




Velocity of Train 

Fig. 61. Relative Velocity. 



enlightenment and the reader must realise for himself the meaning 
of it because mere mental assent to the assertion is useless. We 
will therefore make a very simple model as follows : on a piece 
of tracing paper draw a rectangle ABCD, Fig. 62, and draw a 
horizontal line XY upon a piece of ordinary drawing paper and 
take a number of points 1^, 2r, Sy, 4^, 5^, etc. on this line at 
equal distances apart. Draw also a vertical line ZU and take 
upon it points 1^, 2^, 3^, 4^, 5d also at equal distances apart. 
The poiuts on the line XY represent successive positions of the 
lower right-hand corner of the train window and the pomts on 
the line ZU represent corresponding positions of the rain drop. 
Strictly, the rain drop may be moving with an acceleration so 
that the distances 1^;, 2^ ; 2^, 3^, etc. may progressively increase 
in length, but the whole length ZU is so small that we may 
neglect this refinement. As a matter of fact the resistance to 
the movement of a rain drop makes its acceleration quite small. 



VI] 



RELATIVE VELOCITY 



107 



Now place the rectangle ABCD, representing the window, on 
the tracing paper with the point C at the point ly and mark the 
point 1]) on ZU on the tracing paper ; then shift C to 2y and trace 
the point 2^ as shown in dotted lines on the tracing paper ; then 
move to 3 2- and trace the point Sr and so on. The points on the 
tracing will then join up to an inclined line as shown dotted, and 
this is the direction of the relative velocity between the rain and 
train, i.e. the direction which the rain appears to have from the 
train. Consider any particular point say 4^ ; by the time that 




Fig. 62. Relative Velocity. 

the rain has fallen from 1 j> to 4^) , 4:r has also moved to 4^, and the 
rain strikes 42. in its new position ; to get the apparent position of 
4^^ we set 4:j)4:r back a distance equal to the distance which 4:^ has 
moved in the given time. 

Referring back to Fig. 61 if a6 represents the actual velocity 
of the rain and cb represents the velocity of the train ac will 
represent the relative velocity of the train to the rain. This we 
may regard as a rule which we have proved experimentally; it 
will be found true for any numerical values which may be taken. 

General rule for Relative Velocities. With this preliminary 
explanation we will now give the general rule for relative velocities. 
Suppose that a point A, Fig. 63, is moving with a velocity V4 
with reference to a certain plane in the direction indicated and 
that the point B is simultaneously moving with a velocity v^ 
with reference to the same plane in the direction indicated, 
A and B being positions of the points at the same instant. 

To a convenient scale set out oa parallel to v^ to represent 
Vji in direction and magnitude and to the same scale set out 
ob to represent v^ in direction and magnitude and join ab, then 



108 



VELOCITY CHANGE IN DIRECTION; 



[CH. 



ab is the velocity of B relative to A, i.e. ab is the velocity which 
B appears to have to a person moving with -4 ; it is written 
r^B^. Similarly ba is the velocity of A relative to B. 

It will be noted that this construction is different from that 
employed for finding the resultant of v^ and % ; if the resultant 
had been required we should have drawn ab' to represent v^ as 
indicated in dotted lines and ob' would have given the resultant, 
and is the vector sum. 



6' 




Apparent "^ -^ y, 

direction of A^J^ O'^ 
\^ from B 





V 



Apparent direction 

of B from A ^ 

O 

Fig. 63. Relative Velocities. 

It will be noticed that v^ is the vector sum of Vj^ and r^A* 
i.e. using the vector notation 

Therefore using -< to indicate vector difference we have 

Expressing this in words we see that the, velocity of B relative 
to A is the vector difference between the velocity of B and the velocity 
of A. 

The dotted lines BX and A Y which are each parallel to ab 
are the paths which B and A appear to take from A and B 
respectively. 

Numerical Examples. (1) If a train is runninxj at 30 miles 
an hour, in what direction must a stone be throivn at a velocity of 
60 feet per second to pass in through one open carriage window and 
out through the opposite window ? 

Referring to Fig. 64, the stone must have a velocity relative 
to the train in a direction AB, i.e. at right angles to the direction 
of motion of the train. 

30 miles an hour = 44 feet per sec, so let oa represent 44 feet 
per second; draw a6 at right angles to oa and with o as centre 



VI] 



RELATIVE VELOCITY 



100 



draw an arc of radius representing 60 feet per second cutting 
ab in 6. This determines the point 6 and completes the triangle 
of velocities. We want to find Z aob to obtain the direction in 
which the stone must be thrown. 



/ 



60 feet 
per sec 



'44 feet per site. 



Apparent flight 
of Stone 




44 

Velocity of Tram 

Fig. 64. 

If we draw to scale we shall find that the Z aoh is about 43° ; 
by calculation we have 

, Oa 4:4: ,,^^^ 
cos aob = -f = 7rr{ = /.^o.^. 
ob 60 

From tables we find aob = 42° 50'. 



^ 



(2) A ship A is steaming due N. at a speed of 10 miles an 
hour ; when another boat B is due W. of A and at a distance 
21 miles from it, B starts at a speed of 10 miles an hour in a N.E. 
direction. What is the least distance apart that B ivill attain from 
A and how long after starting will B be at its least distance from A ? 

This question is a little more difiicult, but with the following 
explanation the student should not have much difficulty in 
following it. 

Fig. 65 indicates the position of the boats at the first instant 
under consideration. 

We first draw the vector figure to obtain the relative velocitj" 
ot B to A. Draw ob in a N.E. direction to represent 10 miles 
an hour and oa in a N. direction to represent 10 miles an hour 
also. Then ab is the velocity of B relative to A, i.e. ab is the 
velocity which B appears to have to a person on A. Now draw 
BE parallel to ab; then if ^ were fixed BE would be the path 
taken by the steamer B. The boat B, therefore, will appear from 
A to move along the line BE with a velocity represented by ab. 

If the Aoab be drawn carefully to scale, ab will be found to 
represent 7*65 miles an hour. 



no 



VELOCITY CHANGE IN DIRECTION; 



[CH. 



If we do not plot to scale we can calculate ah as follows; 
Draw ox perpendicular to db. 



Then 



— =sin22J , 
oa ^ 



I.e. 



ax = 10 sin 22|° ; 
/. a6 = 2aa:=20sin22J° 

= 7*654 miles per hour. 




( 






' 7^ 


10 

/ 





Fig. 65. 



Suppose that after a given time, say one hour, B has arrived 
at 2^ in its apparent path BE., then BF will be 7-65 miles and 
AF will be the distance of B from A at that instant ; in other 
words the distances from A to various points on BE give the 
distances apart of the boats at various times. 

The least distance apart of the boats will therefore be given 
by AD, where AD \& drawn perpendicular to BE. By measure- 
ment this should come to 8-04 miles. 

By calculation we have 

AD 

;^-^ = sin22J; 

.-. AD = 21 sin22J = 8036 miles. 



VI] RELATIVE VELOCITY 111 

Now the time taken for this will be the time for B to move 
to D at a speed of 7-65 miles an hour. 

Now BD =19-4 miles (by measurement), 

7? 7) 

(by calculation) j d = ^^^ ^^i 5 

.-. BD = 21 cos 22 J = 19-40 mHes ; 
/. Time= — ^^ -2-53 hours, 
say 2'5 hours. 



SUMMARY OF CHAPTER VI. 

Velocities are combined by the law of vector addition. 

The velocity of B relative to A is the velocity which B would 
appear to have if A were regarded as stationary; it is the vector 
difference between the velocity of B and the velocity of A. 



EXERCISES. VI. 

1. A railway train going at 30 miles an hour is struck by a 
stone moving horizontally at right angles to the train at a velocity 
of 33 feet per second. What are the magnitude and direction of 
the velocity with which the stone appears to meet the train? 

2. A body is moving towards the north at 50 ft. per sec. In 
two sees, afterwards we find that it is moving towards the north- 
east at 60 ft. per sec. Find the magnitude of the added velocity. 

3. A ship is sailing N.E. at 10 miles an hour, and to a passenger 
on board the wind appears to blow from the N. with a velocity of 
14-14 miles. Find the actual velocity and direction of the wind. 

4. Water enters a turbine wheel at an angle of 35° to the cir- 
cimiference, with a velocity of 80 ft. per sec. If the speed of the 
circumference of the wheel is 60 ft. per sec, find the velocity of the 
water relative to the wheel in magnitude and direction. 

5. Two trains each 200 feet long are moving in parallel lines 
with velocities of 20 and 30 miles an hour in the same directions. 
How long will they be in passing? 



112 VELOCITY CHANGE IN DIRECTION [en. vi 

6. Two trains pass one another moving in opposite directions 
on parallel lines of rail, with velocities of 45 and 60 miles per hour. 
The length of one is 420 ft. and of the other 350 ft. How long will 
they be in passing one another? 

7. Two boats each 30 ft. long are rowed at 8 and 7 miles per 
hour respectively, the latter being 80 ft. ahead of the former. Find 
how long before it is bumped; also the time before the former 
draws level with it and the extra time necessary to pass it. 

8. A train is travelling at a rate of 20 miles an hour and a man, 
sitting in a compartment with both windows open, observes a stone 
pass through both windows at right angles to the direction of the 
train. If the stone appears to move 20 feet per second to the man, 
with what velocity must it have been thrown? 

9. A is travelling due N. at a constant speed. When B is 
due W. of A and at a distance of 21 miles from it, B starts travelUng 
N.E. with the same constant speed as A. Determine graphically 
or otherwise the least distance which B will attain from A. 

10. A cyclist is riding due W. at 12 miles an hour and the 
wind is blowing from the S.E. at 5| nriiles an hour. If the cyclist 
carries a small flag, in what direction will this flag fly? At what 
speed would the cyclist have to ride to make the flag fly due N. ? 



CHAPTER VII 

KINETIC ENERGY AND MOMENTUM 

Measurement of Kinetic Energy. We have already explained 
(p. 40) that kinetic energy is the amount of work stored in a 
body in virtue of its velocity but we have not yet explained how 
the kinetic energy can be measured. 

Suppose that a body P, of weight W, starting from rest falls 
from A to J5, Fig. 66, without overcoming any resistance. Then 




Fig. 66. 



if ^ is the vertical distance moved, the weight W has done an 
amount of work upon the body equal to force x distance moved 
by the body in the direction of the force = Wh. Since no work 
has been spent in overcoming resistance, the whole of this work 
must be stored up in the body in the form of kinetic energy 
(K.E.), and since the body was originally at rest and possessed 
no kinetic energy it follows that its kinetic energy at the point 
B is equal to Wh, 
i.e. K.E. = Wh (1). 

A.M. 8 



114 KINETIC ENERGY AND MOMENTUM [ch. 

But we have already shown that for bodies falling freely 
under the action of gravity 

v^ = 2gh, [formula p. 97] 

i.e. h=^-; 

:. we have k.e. = -^ — (2). 

This is a very important formula. In using it, we must note 
that it does not matter how the body has moved in obtaining 
this velocity; all that matters is that the body, somehow or 
other, has attained a velocity v. Then we say that its k.e. is 

2g ' 

We showed for example, on p. 37, that the work done by 
a force depends only on the straight distance in the direction of 
the force between the original or final positions of the body. 
If for instance the body had moved in the irregular path indicated 
in Fig. 66, the work stored in it would still have been Wh and 
therefore the kinetic energy would still be Wh, and since the 
kinetic energy depends Oi.iy on the velocity, by definition it 

must be equal to -^— whatever be the path traversed. 

The direction of the velocity will be different in the two cases, 
but that does not matter so far as kinetic energy is concerned. 

Change in kinetic energy. Suppose that a body of weight W 
has at one instant a velocity u and at some subsequent instant 
it has a velocity v. 

Then its kinetic energy has changed from -^ — to -^ — . 

2gr 2g 

W 
:. Change in k.e. = ^~ {v^ - u^) (3). 

Complete energy equation. We have shown on p. 40 that 
work cannot be destroyed and that the difference between the 
amounts of work done by the effort and the resistance must be 
equal to increase or decrease of the kinetic energy. 

We therefore have 

Work done by effort = Work done against resistance 

+ Gain in kinetic energy, 
i.e. Pjg = Eli + K.E. gained (4). 



vii] KINETIC ENERGY AND MOMENTUM 115 

One of the best examples in practice of the use of kinetic 
energy arises in the use of flywheels to steady the motion of 
machines. We shall deal with flywheels later under rotating 
bodies. 

Numerical Examples on Kinetic Energy. (1) A bullet 

weighing 3 ounces is discharged from a rifle with a velocity of 1200 

feet per second. How much kinetic energy does it possess and how 

far will it be able to move a body the resistance to whose motion is 

10 tons if we neglect the energy lost in the impact ? 

3 
The weight of the body in lbs. = ^ = f^ 5 

_Wv^_ 3^ 1200 X 1200 
/. K.E. - 2^ - iQ^ 32-2 

= 83,800 ft.-lbs. nearly. 



/ 



If the resistance to the motion of a body is i? lbs., the work 
done in moving the body a distance s feet in opposition to the 
resistance is equal to Bs ft.-lbs. 

In our present case B = 10 tons = 22,400 lbs. 

/. 22,4005 = 83,800, 

83,800 or,. X . 

We wish to warn the student that the above calculation is 
chiefly of academic interest and of value as an exercise in applying 
the formulae. As a matter of fact considerable energy is absorbed 
in the impact, being converted into the thermal form of energy, 
and the resisting force will not be constant. 

(2) A train weighing 100 tons gets up a speed of 30 miles an 
hour in 1 mile from rest on the level, the air and other resistances 
being equivalent to a force of f ton. What constant tractive effort 
is required ? 

In this case original k.e. =^ 0. 

Wv"^ 
After one mile k.e. = -^^^ — , 

2g 

V = 30 miles an hour = 44 feet per sec, 

100 X 2240 X 44 X 44 



K.E. = 



2 X 32-2 
= 6,734,000 ft.-lbs. nearly. 

8—2 



116 KINETIC ENERGY AND MOMENTUM [ch. 

The distance travelled in getting up a speed of 44 feet per 

second is 1 mile, i.e. 5280 feet. 

Therefore if the effort is F lbs. we have 

Work done by effort = Work against resistance + Gain in k.e. 

T^ •. 3, 3x2240„ Tconiu 

Resistance = -r ton = :: lbs. = 1680 lbs. 

4 4 

.-. F X 5280 = 1680 x 5280 + 6,734,000 

or (F - 1680) 5280 = 6,734,000, 

6,734,00 

^-^^^^=-5280"- 

= 1275 lbs. 

.-. F = 1275 + 1680 = 2955 l bs. 

(3) Taking the numerical example worked on p. 48, find the 
maximum velocity and the velocity at the end if the initial velocity 
was 10 feet per second and the body weighs 1000 lbs. 

We have given that the gain in k.e. up to the point of 

maximum velocity = 2450 ft. -lbs. 

^ .^. , Wu'' 1000 X 10 X 10 __.. ,, 

Initial K.E. = -J- = — 2ir32^~~ ^ ft.-lbs. ; 

.*. K.E. at maximum velocity = 1553 + 2450 ft.-lbs. = 4003; 
.*. if V is the maximum velocity 

i|^=4003, 

lOOOi;^ ..., 

2T32^ = ^^^^' 

2 2 X 32-2 X 4003 ^._ . 

^ '^ looo ^ ^^^^ ' 

.-. V = \/258 = 16 ft. per sec. 

We next find the velocity at the end of the motion as follows. 
At the end, the excess work which appears as kinetic energy 
was found to be 1000 ft.-lbs. 

Therefore k.e. at end -= k.e. at beginning + work added 

= 1553 + 1000 

= 2553 ft.-lbs. 

:. »2 = 2553, 

2 _ 2gr X 255^ ^ 2 x^-2 x^553 
•• ^ ~ 1000 ~ 1000 

= 164 nearly ; 
/. V = \/i64 = 12-8 feet per sec. 



VII] KINETIC ENERGY AND MOMENTUM 117 

The connection between Force and Acceleration. Suppose 
that a force F is acting upon a body weight W . It has been 
found experimentally that the body will be given a uniform 
acceleration a in the direction of the force and that a bears the 
same ratio to g the gravity acceleration as F bears to W. 

We therefore have the rule 

a F 

-g-W (^)- 

F-^ (6). 

This law is one of the most important in the whole range of 
mechanics and the student must master it before he can hope to 
appreciate the interest and importance of the subject. It was 
discovered by Newton and is often called the second law of 
motion although it is usually expressed in different language. 
We have already considered Newton's other two laws of motion 
and will summarise them again a little later. For the present 
we will endeavour to become familiar with these formulae. 

Suppose that a body of weight W is moved by a force F a 
very short distance s in the direction of F and that its velocity 
at the beguining of the distance is u and at the end is v. 

Then work done = F . s. 

If this all goes in increasing the k.e. we have 

F . s = gain in k.e. 

==|(^'-«') (7). 

Now if s is so short that the force F is constant over it and 
that the acceleration is also constant, we have by formula (7), p. 96, 

v^ = u^ + 2as, 
i.e. v^ — u^ = 2as. 



Putting this in (7) 



W 

F.s = ^.2as, 



i.e. :.F=^^ (S). 

9 

We can thus deduce the result from the principle of the 
conservation of energy. 



118 KINETIC ENERGY AND MOMENTUM [en. 

Numerical Examples. (1) The piston of an engine weighing 
20 lbs. is given a retardation of 6 feet per second per second. What 
backward pressure will be acting on the piston ? 

We shall discuss the crank and connecting-rod mechanism of 
an ordinary engine in a later chapter (p. 258) and those students 
who would like to understand this question with special reference 
to its application are recommended to refer to that description. 
Putting our values in 

P _ ^^^ _ 20 X 6 ( — indicates back- 
er 322 ward pressure) 
= 3-73 lbs. Ans. 

(2) A train zveighing 100 tons gets up a speed of 30 miles an 
hour in 1 mile from rest on the level, the air and other resistances 
being equivalent to a force of | ton. What constant tractive effort 
is required ? 

We have already worked this example on p. 115 from the 
work point of view ; now let us work it from that of the accelera- 
tion. 

We have ^2 = ^2 + 2as, 

t; = 44 feet per sec, w = 0, s = 5280 feet, 

.-. 44 X 44 = 10560a, 

44 X 44 - 00 X 4. 
a = = 183 feet per sec. ; 

. T, u ^ X Wa 100 X 2240 x -183 

.'. Resultant force = - — = irr—r 

g 32-2 

= 1274 lbs. 
Now resultant force = effort — resistance, 
i.e. 1274 = 2?^- 1680; 

.-. F = tractive effort = 2954 lbs. Ans. 

We think that as a general rule the student will ^nd the work 
method of solving problems easier to deal with than the accelera- 
tion method, but it is somewhat a matter of individual taste. 

(3) Take example (2) and find the effort required to give the 
same speed in the same distance up an incline of 1 in 100. 

Solution (i). By acceleration. 

In this case there is in addition to the air and like resistances 
a resistance equal to the resolved component of the weight of 
the train in the direction of the tractive effort. 



VII] 



KINETIC ENERGY AND MOMENTUM 



119 



This is an example of the inclined plane and we proved (p. 58) 
that adb, Fig. 67, is the triangle of forces. 

.'. Component down plane = -y — = 1 ton nearly 

= 2240 lbs. 
/. Resistance now = 2240 + 1680 = 3920 lbs. 
Resultant force up plane = 1275 lbs. (as before). 
.♦. Tractive effort = 3920 + 1275 
= 5195 lbs. 




100 
r 100 Tons 

Fig. 67. 

Solution (ii). By work equation. 

When the train goes 1 mile its weight is lifted by an amount 
very nearly equal to yj^ mile =52-8 feet. 

.*. Work done = 100 x 2240 x 52-8 ft.-lbs. against gravity 
= 5280x2240; 
;. Work done by effort 

= Work done against resistance + gain in k.e., 
F X 5280 = 1680 x 5280 + 2240 x 5280 + 6,734,000, 
.-. 5280 {F - (1680 + 2240)} = 6,734,000, 

^_ 3920 = 5.^^^5^=1275 lbs., 



i.e 



I.e. 



5280 
F = 1275 + 3920 = 5195 lbs. 



Momentum. We have seen that when a body is moving 

with a certain velocity, it has a certain amount of kinetic energy. 

It is said also to possess momentum, the amount of momentum 

being defined as follows. If a body of weight W lbs. is moving 

with a velocity v feet per second, its momentum is equal to 

Wv Wv 

— , i.e. ^^^ in lb. -ft. -second units. 

Momentum is a vector quantity because it has direction as 
well as magnitude. 

We can therefore combine momenta by the law of vector 
addition. Change of momentum is found in exactly the same 
manner as change of velocity (p. 104). 



^ 



120 



KINETIC ENERGY AND MOMENTUM 



[CH. 



Numerical Example. What is the momentum possessed by 
the bullet referred to in example (1) on p. 115.? 

3 

Wt. = 3 oz. = Y« lbs., V = 1200 feet per second ; 



16 

.*. Momentum = 



3 X 1200 



7 units. 



16 X 32-2 

Dimensional equations. We can find the dimensions of the 
units in any formula by writing a dimensional equation as follows : 

ft. 
Wv 
9 



Momentum = 



lbs. X 
IT 



sec. 



sec. 2 

_ lbs. X sec.H x^ 
^. x ^eq^. 

= lbs. X seconds. 

In these dimensional equations we cancel out dimensions 
according to ordinary rules of fractions, and students should 
write such an equation whenever they are not certain as to whether 
or not their formula is in the right dimensions. Mere numerical 
coefficients are not counted. 

As another example take kinetic energy : 

ft. \2 

K.E. = 



\sec./ 



2g ft. 

sec. 2 

_ lbs. X ft.H X s^sp.2 
~ §eq.2 X ^. 
= lbs. X ft. 

We know that energy should be in work units and the fact 
that this comes to lbs. x ft., i.e. in work units, shows us that the 
formula is of the right order. 

The importance of Acceleration in Traction Problems. 
In all branches of traction engineering — railways, tranways and 
motor cars — the question of acceleration is of the greatest possible 
importance and students who wish to specialise later in any of 
these branches should make themselves quite clear on this subject 
of the connection between effort or force and acceleration. 



VII] KINETIC ENERGY AND MOMENTUM 121 

Electric traction is supplanting steam traction for suburban 
traffic, not because it is less expensive but because the electric 
trains can get up speed more quickly or in other words they can 
have a greater acceleration. If we have a fixed resultant effort 
F available, the acceleration a is given in terms of the weight 
W to be moved by the relation 

and the smaller we make the weight W, the larger the acceleration 
a will become. This is why we want to keep down the weight 
as much as possible if we want to start quickly ; at the same time 
the weight must be enough to get sufficient grip upon the rail 
or road to prevent the driving wheels from slipping; this is 
allowed for in electric trains by having motors on several of the 
carriages. If we have two vehicles such as bicycles exactly 
similar but very different in weight, say one made of aluminium 
and one of steel, we speak of the heavier one as more difficult to 
push even on the level; we mean really that it is more difficult 
to accelerate or start. There is practically no difference in the 
force required to keep the heavy and the light one moving at 
a given speed once they have been started. Modern traction is 
principally concerned with the question of getting up speed — 
and, as far as brake problems go, of slowing down — and to deal 
with these problems we must know how to calculate the accelera- 
tion when we know the resultant effort and the weight of the 
body. 



SUMMARY OF CHAPTER VII. 

Kinetic Energy (k.e. ) = -^ — . 

W 

Change in k.e. = ^ {v^ — u^). 

Work done by effort = Work done against resistance + Gain in 
kinetic energy. 

9 

Wv 

Momentum = . 

9 



122 KINETIC ENERGY AND MOMENTUM [ch. 



EXERCISES. VII. 

1. A body of weight 10 lbs. moves with a linear velocity of 
800 ft. per min. Find its kinetic energy in ft. -lbs. 

2. A train weighing 150 tons is running on a level road at 
30 miles per hour. The resistances are equal to 12 lbs. per ton. 
If steam be shut off how far will the train run before coming to 
rest ? Give answer in yards. 

3. A body weighing 100 lbs. increases its velocity from 25 to 
35 yds. per sec. Find the increase in its kinetic energy. 

4. In a Fly Press the radius at which the balls revolve (there 
are two each weighing 12 lbs.) is 10 ins. and the number of revolutions 
per min. is 150. If the die be brought to rest after stamping through 
a piece of metal | in. thick, what is the average force resisting the 
blow? 

5. A shot weighing 6 lbs. leaves the mouth of a gun with a 
velocity of 1000ft. per sec; determine the number of ft. -lbs. of 
energy accumulated in it and the mean pressure exerted by the 
exploded powder behind it if the length of the bore is 5 ft. 

6. A body weighing 108 lbs. is placed on a smooth horizontal 
plane, and under the action of a certain force describes from rest 
a distance of Hi ft. in 5 sees. Find the force in lbs. 

7. The rim of a flywheel weighs 9 tons and its mean linear 
velocity is 40ft. per sec; how many ft. -tons of work are stored 
up in it? If it is required to store the additional work of 9 ft. -tons 
what should be the increase in velocity? 

8. A train weighing 50 tons is impelled along a horizontal 
road by a constant force of 550 lbs. ; the frictional resistance is 
8 lbs. per ton; what velocity will it have after moving from rest 
for 10 mins., and what distance will it describe in that time? 

9. A car weighing 2^ tons and carrying 40 passengers of average 
weight 145 lbs. each is travelling on a level rail at 6 miles per hour. 
What is the momentum? 

10. In the previous example what average force must be 
exerted to bring the car to rest in 2 seconds, and if that force is 
constant what distance will the car travel before it comes to rest ? 

11. A ship weighing 2500 tons is propelled at 20 knots (1 knot = 
6080 ft. per hour) by engines of 8000 h.p. Estimate the distance 
which will be traversed by the ship whilst an amount of energy 
is developed by the engines equal to the kinetic energy of the ship. 



VII] KINETIC ENERGY AND MOMENTUM 123 

12. A weight of 50 lbs. is moving at a speed of 15 feet per second 
and it is acted upon for 20 seconds by a force of 20 lbs. in the direction 
of motion. What is the distance moved through during the time? 

13. A train weighing 250 tons is moving at 40 miles per hour 
and is stopped in 10 seconds. What is the average force causing 
stoppage ? 

14. A planing machine table weighs 2 tons and has a retardation 
at the end of its stroke of 3 feet per second per second. What 
thrust will this cause on the driving mechanism ? 

15. When starting, a locomotive exerts a tractive force of 
4 tons upon a train weighing 200 tons. Calculate the acceleration 
(neglecting friction), and the velocity after 1 minute. 

16. A piston and rod and cross-head weigh 330 lbs. At a 
certain instant, when the resultant total force due to steam pressure 
is 3 tons, the piston has an acceleration of 370 feet per second per 
second in the same direction. What is the actual force acting on 
the cross -head ? 



CHAPTER VIII 

Newton's laws of motion: impact 

We will now consider collectively Newton's Laws of motion 
which are the foundation of the whole scientific treatment of 
mechanics. 

They may be enunciated as follows : 

1. A body continues in a state of rest or uniform motion in 
a straight line unless it be acted upon by some external force. 

2. The rate of change of momentum is proportional to the force 
applied and takes place in the direction of the force. 

3. To every action there is an equal and opposite reaction. 

1. The first law is sometimes called the law of inertia ; inertia 
bemg the property of a body which resists a change in its state 
of rest or motion. It follows from this law that if there is a 
resultant force acting upon a body, it must either change its 
velocity if it is already moving or else start moving if it is 
stationary ; in either case the body will be given an acceleration, 
which may be negative, i.e. a retardation. In all engineering 
problems dealing with bodies which from their very nature must 
be stationary — e.g. structures such as bridges, roofs, dams, etc. — 
we know from this law that all the forces acting must neutrahse 
each other, or in the language of mechanics their resultant must 
be zero. 

This law cannot be rigorously demonstrated experimentally 
because it is impossible for us to move bodies without external 
forces being brought into play; we have referred to these as 
passive resistances. A stone thrown along a road soon comes to 
rest on account of the forces — called frictional forces — caused by 
the roughness of the road, but if thrown along a surface of ice 
which has very little friction the stone will run for a very long 



CH.vni] NEWTON'S LAWS OF MOTION: IMPACT 125 

way. These frictional forces are the bugbear of the engineer; 
he has for centuries been trying to make tiiem as small as possible 
but he can never get rid of them altogether. If he could, the 
world would be a very different place. As a matter of fact 
frictional resistances, which we will deal with in detail later, are 
of great value in certain cases. In frosty weather for instance 
we throw sand down upon the roads to increase the friction 
because without it we should not be able to get sufficient grip 
to propel our vehicles. What we should like to be able to do 
is to bring frictional forces into play when they are useful and 
eradicate them when they are not, but natural phenomena will 
not change for our convenience; all that we can do is to study 
them as closely as possible in order to use them to our greatest 
possible advantage. 

2. This is Newton's way of expressing the law that we have 

Wa 
reduced to symbols in the form F = for the case in which 

the weight does not change. We will consider that case in detail. 
Suppose that in a very short time t seconds the velocity of a body 
of weight W changes from v feet per second to v' feet per second. 

Wv Wv W 

Its change of momentum is equal to = — (v — Vi) and 

^ g g g ^' 

this takes place in a time t. 

:. Change of momentum per second 

= Rate of change of momentum 

g ^ g t 

— . — -] is the rate of change of velocity and this we 

have called the acceleration (a), so that we have 

Wa 
Rate of change of momentum = - . 

g 

Newton's law states that the rate of change of momentum is 
proportional to the force applied; whereas in our formula we 
make it equal to the force applied. This is because we choose our 
units of force and momentum so that the proportionality becomes 
an equality. 

3. This law is very important and will become more clear if 
we give some explanatory considerations. 



126 NEWTON'S LAWS OF MOTION: IMPACT [cH. 



I 
I 




Take the case of a weight hung on the end of a rod, Fig. 68 ; 
the result of that action will be to cause the rod to stretch. 
The amount of stretching will be very small but it can be 
measured by delicate instruments called "extensometers." This 
stretching brings into play forces between the molecules of the 
rod tending to resist the motion. These molecular forces are 
called stresses and act across every section that we consider. 
The stresses increase with the 
amount of the stretching, which 
will continue until the resultant 
of the stresses is equal to the 
weight W; this resultant of the 
stresses is the reaction which is 
equal and opposite to the action, 
i.e. the weight. If the reaction 
contributed by the stresses is not 
equal to the weight, there will 
be a resultant force acting upon 
the rod below the section under 
consideration. From the first 
law this must cause the change 
of state of rest of the rod, i.e. 
must start it moving. This is 
exactly what happens when the 
load is so great that the rod 
breaks. For every material there 
is a certain maximum stress that 
it is capable of calling into play 
so that the resultant stress can 
never be more than a certain 
amount; if therefore the weight 
is greater than this, motion must 
take place and the rod fractures. When the load is removed 
the rod returns to its original length (unless the material is not 
perfectly elastic*), the return movement showing the existence 
of the stresses. 

As another example take the case of a man striking his fist 
against a wall. The wall presses just as hard on the man's hand 




4 W 



Fig. 68. 



• See p. 139. 



vni] NEWTON^S LAWS OF MOTION: IMPACT 127 

as his hand presses on the wall and the feeling of pain which the 
man experiences is a proof of the existence of the reaction. 

Next take the case of a traction engine pulling along a truck. 
The truck pulls back on the engine just as much as the engine 
pulls on the truck. How then is it that the truck goes along 
at all ? We may answer that the truck does not move relatively 
to the engine and that if there were not equality between the 
forces between them there would be a resultant which would 
cause relative motion. 

At the same time it is true that the engine must exert a 
greater effort than the force with which it pulls the truck. 
Referring to Fig. 69 let F be the effort which the engine exerts 



Traction Engine 




upon the ground ; it is only by means of the grip upon the ground 
that the engine can pull ; that is why the driving or back wheels 
are usually roughened. A traction engine would be absolutely 
useless upon ice because the wheels would merely slip round. 
If you watch the locomotive of a heavy train start you will 
usually notice that the driving wheels will slip and buzz round ; 
the driver then operates a device for projecting sand under the 
driving wheels to increase the grip. The ground has to be able 
to exert the same effort F as a reaction or else, as we have seen, 
slip occurs. 

Now part of this force will be spent in overcoming the resist- 
ances in the engine and accelerating it and part in overcoming 
the resistances in the truck and accelerating it. The force P 
therefore which has to be transmitted across the coupling is less 
than the force F which the engine has to exert upon the ground. 
One form of reply therefore to the question as to how it is 



128 NEWTON'S LAWS OF MOTION: IMPACT [ch. 

that the engine can pnll the truck is that the engine only exerts 
the same force upon the truck as the truck exerts upon the engine 
but that the engine exerts upon the ground a greater force than 
the resistance to the motion of the truck. 

Let us consider this problem in somewhat greater detail. 

Let Re and jB,. be the resistances to motion of the engine and 
truck respectively and let We and Wt be their weights. These 
resistances will depend upon the weights and the velocity to 
some extent and can of course only be found accurately by 
experimental determination for any given vehicle. 

We then have 

Total effort = Effort spent in overcoming resistance 

4- Effort used in accelerating, 

i.e. j^ = i?, + iJ, + Z-^ + iL^^. 

g g 

Of this P the force transmitted through the coupling is equal to 



Rt + 



\/ 



g 



Numerical Example. A man weighing 12 stone is going up 
in a lift which has an acceleration of 3 feet per second ^^er second ; 
what pressure does he exert on the floor ? 

In this case the floor of the lift is exerting sufficient force to 
lift the man and in addition give him an acceleration of 3 ft. per 
sec. 2 

Force required for acceleration = 

12x 14x3„ 
= 32-2 ^^^' 

= 15-6 lbs. nearly. 

Therefore total upward pressure = 12 x 14+ 15*6 

= 183-6 lbs. 

Since the pressure of the floor upon the man must be exactly 
equal to his pressure on the floor, the man must exert a pressure 
of 183-6 lbs. 

As the lift slows down, the acceleration is negative so that 
the pressure is less than the man's weight. The same occurs 
when the lift starts downwards and accounts for the unpleasant 
feeling which often accompanies a quick-stopping lift. 



vm] NEWTON'S LAWS OF MOTION: IMPACT 129 

Impact and Impulse. Up to the present we have considered 
only the cases in which the forces acting are gradually applied 
and act over a considerable length of time. In some cases, 
however, the forces act over an extremely short time, as in an 
explosion or the blow of a hammer, and then we require to be 
able to estimate the force of the blow. Such suddenly applied 
forces are called impulsive forces and it is very difficult to calculate 
accurately the maximum force produced by a blow. AU that 
we can do is to find the average value of the force if we know the 
short time during which the force or impulse acts. It is very 
necessary to realise the difference between the average force and 
the maximum. The determination of the maximum force of an 
impact is a very troublesome and advanced problem, but a 
knowledge of the average force is of considerable value to us in 
some problems. 

Suppose that a body of weight W lbs. moving with a velocity 

V feet per second is suddenly brought to rest in t seconds. The 

average force F produced by the blow will be equal to the rate 

of change of momentum, i.e. to the momentum destroyed per 

Wv 
second. But the original momentum was — and it was destroyed 

Wv . 
in t seconds, so that — — is the momentum destroyed per second. 

tf 

Therefore we have average force produced 

Wv 

= ^=^ lbs (1). 

gt 

Numerical Example. A hammer weighing 2 lbs. and having 
a velocity of 30 feet per second strikes a blow lasting y^^ second. 
What force is produced by the blow ? 

Substituting directly in formula (1) we have 

32-2x4 ^-2 

This question of impact is of very great importance and 
through loose use of language many people have wrong ideas 
about it. For instance a man may ask what force he can exert 
with a certain hammer; the correct answer is that you cannot 
tell because you must know the time that the blow lasts before 
you can know the force of the impact. If the hammer loses its 

A.M. 9 



/ 



130 NEWTON'S LAWS OF MOTION: IMPACT [oh. 

momentum very quickly it strikes a heavy blow, but if it takes 
comparatively long to do so only a light blow results. A certain 
hammer will strike a very much heavier blow upon cast iron 
than it will upon wood and it will strike a much harder blow 
upon wood than upon sand. A carpenter does not use a heavy 
hammer for a chisel because wood is soft and does not require 
a very heavy force to cut it, the tool therefore has a wooden 
handle to absorb some of the shock and make the cut longer; 
but when tooling a very hard material a greater force is required 
so that as "snappy" a blow as possible is delivered. When we 
wish to avoid percussive forces we provide some device such as 
springs for making the force act over a longer time; for this 
reason we hang our carriages upon springs to deaden the effect 
of impulsive forces. 

Equality of Momentum before and after impact. If two 
bodies collide or an explosion occurs between them, there are 
forces acting between them and by Newton's third law the forces 
are equal and opposite. Therefore the rates of change of 
momentum of the two bodies are equal and opposite. Thus one 
body gains in momentum in any direction as much as the other 
loses or the total momentum of the two bodies in any direction 
remains unchanged. This is sometimes called the law of the 
conservation of momentum and is an extremely important principle. 
Momentum cannot be destroyed by impact ; its effects can only 
be got rid of by swamping it, i.e. hy transferring it to a body 
whose weight is so great that the resulting change in velocity is 
negligible. If for instance a stone hits the ground, it loses its 
own velocity and therefore its momentum but it communicates 
the momentum to the earth. The weight of the earth is, how- 
ever, so immense that the resulting velocity is so small that for 
all practical purposes it may be taken as nothing. We have 
constant examples of the reactive forces caused by a sudden 
change of momentum ; we will consider some of them in detail. 

Impact of bodies. Referring to Fig. 70, let one body of 
weight W possessing a velocity u collide with another body of 
weight Wi moving in the same direction with a velocity w^. 
Then the total momentum before impact 

-a 171 = ^ ^-J^ (2). 

g g 



vni] NEWTON'S LAWS OF MOTION: IMPACT 131 

If the velocities after impact are respectively v and Vi in the 
same direction we shall have that after impact 

m = ^%-^ (3). 

g 9 

Since these must be the same we have, cancelling out g which 
is common throughout, 

Wu + W^Ui = Wv+ W^Vi (4). 

This equation alone is not sufficient to determine the velocities 
after impact unless the bodies are "inelastic," i.e. they do not 
rebound. The accurate treatment of the impact of elastic bodies 
such as billiard balls is very difficult and beyond the scope of the 
present book. 




^^^^5 



m//m 



Fig. 70. 

Restricting therefore our consideration to bodies which do 
not rebound we shall have the two bodies going on together at 
the same velocity after impact; if this common velocity is v we 
shall have in equation (4) 

Wu + W^Ui = Wv -{- W^v 
= (W+ Wi) V, 



••^"V w + w^ J ^^^' 



Loss of energy at impact. Although there is no loss of 
momentum at impact there will always be a loss of kinetic 
energy. This loss of energy is evidenced by the noise produced 
by the impact and also by the heat produced ; bullets for instance 
become very hot on impinging against anything. 

The loss of energy can be expressed in formulae as follows: 
taking as before the case in which there is no rebound, we have 

Wu'^ W u ^ 
Total Kinetic Energ}^ before impact = — ! ^—^ • • (6), 



Total Kinetic Energy after impact = '^ ^ — — — 

= (from(5))i-^^f^^±^V 
^ ^ ^' 2g \ W-^W^ J 



(7) 



^ {Wu + W,u,)^ 

2g{W + Wi) ^^^• 



• —2 



132 NEWTON'S LAWS OF MOTION: IMPACT [ch. 

.*. subtracting (8) from (6) and multiplying out, we have 
Loss of K.E. 

- 2sr + 2g' 2g(W+W^) 



2g[ (Pf+IFi) 

1 \ WWiU^+WWiUi^-'2WWiUUi ^ 



{uUi^-Wj^Uj^} 



2g{W+Wi) 



{u^ — 2uui — i^i^j 



Now {u — Ui) is the relative velocity between the two bodies 

so that we have 

WW 
Loss of K.E. = ^ ,T^ — ^=-r X square of relative velocity. 
2g{W+Wi) 

It should be noted that if one of the bodies is moving in 
a direction opposite to that of the other, one of the velocities 
should be considered negative. 

This principle is useful in questions dealing with the waste 
of energy due to a sudden contraction in a water pipe. 

Numerical Example. Water is flowing through a pipe and Jias 
a velocity of Sfeet per second until it meets a sudden contraction 
when the velocity is suddenly increased to 6 feet per second. How 
much kinetic energy is lost per pound of water flowing ? 

In this case the change of velocity is not absolutely instan- 
taneous in practice because the water will curl round somewhat 
as shown in Fig. 71, but experiments have shown that the present 
method of treatment based upon impact formulae gives results 
that are approximately correct. 

Now the velocity increases and therefore the kinetic energy 
increases; the pressure, however, of the water will diminish. 
If the change of section were gradual this diminution of pressure 
would be such as to keep the total energy constant, but with 
the abrupt change there will be loss of energy and the pressure 
will therefore be less still. 

In our example Tf = 1 lb. and W^ is very large, because the 



vra] NEWTON'S LAWS OF MOTION: IMPACT 133 

pipe must be fixed to something very heavy else it would be 
pushed along. 

I X W 
:. Energy lost per pound = ^^ ^ ^ (3 - 6)". 




1 



JT 



I) 



Kg. 71. 

w 

Now , ^ = 1 to all practical purposes if Wi is very great ; 
if for instance Wi = 10,000 lbs., 

Wi 10,000 , 
1 + TFi "^ 10,001 • 

/. Energy lost = ^^^^ x (- 3)^ 

= ^ =-14ft.-lbs. 
64*4 — — — 

Recoil of guns, etc. Everyone who has used a rifle knows 
that it recoils or kicks back as the shot is fired. This is because 
before the explosion the shot had no velocity and therefore no 
momentum; it is suddenly given a velocity and momentum at 
the explosion, and since the total momentum before and after 
the impulse must be zero, the rifle is given an equal and opposite 
momentum which will drive the rifle backwards to an appreciable 
extent if it is not securely held. 

An exactly similar effect is noticeable in the case of hose 
pipes. Firemen have to hold the hose pipe quite firmly, otherwise 
the pipe would jump backwards. This recoil was made use of 
in the simplest and earliest form of steam-engine (Hero's engine) 
and in an early form of water wheel known as "Barker's Mill," 
the modem form of which is the sprinkler used to water lawns. 
In this form the water rushes out and drives the radiating arms 
backwards. 

Experiment. A very simple and instructive form of Hero's engine 
can be made as follows: Take a piece of glass tubing such as is used very 
largely in chemical apparatus and by heating in a Bunsen burner and drawing 



134 NEWTON'S LAWS OF MOTION: IMPACT [ch. 



down obtain a piece about ^ inch long with a thin neck at each end, as shown 
in Fig. 72 (a). Now cut ofiE one end as indicated in dotted lines and close this 
end by heating in the flame and then blow a bulb at the end as sho^vn at (6). 
Next soften the glass just above the bulb at A and bend the stem over to 
resemble a glass retort and then bend over at right angles to the plane of the 
paper at JB, about half an inch from the top. Our reaction steam turbine is 
now complete. 



B 



(a) 



0>) 



A-- l-i--^::- 





/ 



XL- 



Fig. 72. 

We have next to get some water into it. To do this warm it carefully 
without making it very hot and thus drive out some of the air from it; 
then dip the open end into water and the bulb will fill partly as the air cools. 
Now hang it up as indicated in (c) by means of a piece of thread, or better by 
means of a stump of wire joined to a piece of thread, and hold it over a gas 
flame, taking care not to bum the thread. The water will then boil and the 
steam rushes out at the end and makes the ''engine" buzz round in merry 
fashion. 

This question of recoil is of very great importance in the case 
of large guns, particularly those mounted on ships. In the design 
of battleships great care has to be taken that the stability is 
sufficient to bear the tremendous backward thrust of a broadside. 



v/?n?y/'\ 



w^^m^A 



V <r 



^^ZZZZZZZZZZZZZSZ 



■^^^w 



''^mmm^ 




Fig. 73. 

Referring to Fig. 73, let W be the weight of the shell and 
If 1 be the weight of the gun and let v be the velocity with which 



VIII] NEWTON'S LAWS OF MOTION: IMPACT 135 

the sheU is driven forward as a result of the explosion ; then the 
gun will be driven backward with a velocity v^. Then if we 
neglect the momentum of the gases resulting from the explosion, 
the backward momentum of the gun must be equal to the forward 
momentum of the shell, 

W.v, Wv 
i.e. — ^— ^ = — , 

9 9 

Wv 

Numerical Example. A gun weighing 40 tons fires a shell 
weighing 100 lbs. vnth a velocity of 1500 feet per second. What is 
the velocity of recoil ? 

In this case W^ = 40 tons = 40 x 2240 lbs., 
W = 100 lbs., 
V = 1500 feet per second ; 
Wv 100 X 1500 , ^ , 

^ Wi 40 X 2240 ^ 

= 1*67 feet per second. 

The action of a pile-driver. The operation of driving a pile 
into mud or soft earth gives us a familiar engineering application 
of the principles of impact. The pile A, Fig. 74, of wood, or 
nowadays of reinforced concrete, is driven into the mud by blows 
with a hammer ' ' tup " or * ' monkey "5. This tup has in one form 
of pile-driver a hook C which is weighted so as normally to engage 
an endless chain D which moves upwards and carries the tup with 
it. The hook then meets a releasing or "trip" device E which is 
suspended by a rope so as to be adjustable in height; the tup is 
then freed from the chain and drops on to the pile, thus driving 
it in to an extent dependent upon the resistance of the mud or soil. 

Let W be the weight of the pile and TFj that of the tup and 
let h be the height through which the latter falls. 

Then its velocity v^ is given by v^^ = 2ghy 

or Vi = J2gh ; 

therefore if the tup does not rebound we shall have that if v is 

the velocity with which the pile and tup move, 

Momentum after impact = (TF + Wi) v, 

Momentum before impact = WiV^ ; 

•• ^~ W+ TFi^^i- 



/ 



136 NEWTON'S LAWS OF MOTION: IMPACT [ch. 



If the resistance to the pile were uniform and equal to B we 
should have that if 5 is the short distance moved, 

R .s = Work done 

s= K.E. possessed just after impact 
by tup and pile 

2g.{W+Wi)^ ~W+Wi 



(^/')' 



*»■"> — "r^ jy ^ • • • v^;* 

s 



i'-w) 



This formula is not, however, 
strictly applicable to this problem 
because the resistance is not uni- 
form. 

A formula which is used a good 
deal in practice in order to determine 
the safe load P to put upon a pile is 

2W^h 



P = 



(2), 



x+ 1 

where h is the drop of the tup in 

feet, 
and X is the penetration in inches 

of the last blow. 
This formula professes to give a 
safe load on the pile equal to J of 
the average resistance of the last 
blow. 

That is P = 77 of our formula (1). 

Putting therefore R = 6P, 

p ^ W,h 




we have 



6s 






Fig. 74. Pile -Driver. 



vin] NEWTON'S LAWS OF MOTION: IMPACT 137 

or if ^ is in feet and for s which is also in feet we write y^ where 
x is in inches we shall have 



P = 



2 V ^ Wj 



2W^h 



2W^h 



(3). 



w 

If, therefore, .^ . a; is equal to 1, the formula used in practice 

is equivalent to that which we have obtained from theoretical 
considerations which are not strictly applicable. 



SUMMARY OF CHAPTER VIH. 

Newton's Laws of Motion. 

(1) A body continues in a state of rest or uniform motion in 
a straight Una unless it be acted upon by some external force. 

(2) The rate of change of momentum is proportional to the 
force applied and takes place in the direction of the force. 

(3) To every action there is an equal and opposite reaction. 

Suddenly applied forces are called impulsive forces. 

gt 

Momentum before and after impact is equal. 

Although no momentum is lost in impact there is always a loss 
of energy. 



138 NEWTON'S LAWS OF MOTION: IMPACT [ch. viii 



EXERCISES. VIII. 

1. A hammer head weighing 2^ lbs. and moving with a velocity 
of 50 ft. per sec. is stopped in -001 second. What is the average 
force of the blow? 

2. A ship weighing 2000 tons and moving with a speed of 

3 knots is stopped in 1 minute. Neglecting the motion of the 
water find the average retarding force if 1 knot is 6080 feet per 
hour. 

3. A cage weighing 1000 lbs. is being lowered down a mine 
by a cable. Find the tension in the cable (1) when the speed is 
increasing at the rate of 5 feet per second per second; (2) when 
the speed is uniform; (3) when the speed is diminishing at the rate 
of 6 feet per second per second. The weight of the cable itself 
may be neglected. 

4. A jet of water 1 inch in diameter falling from a height of 
200 feet strikes a fixed hemispherical cup so as to reverse its direction. 
Find the force which it exerts upon the cup assuming that the 
jet has 90 per cent, of the full velocity due to its height of fall. 

5. A gun delivers 400 bullets per minute, each weighing '6 oz., 
with 2000 feet per second horizontal velocity ; neglecting the momen- 
tum of the gases, what is the average force exerted upon the gun ? 

6. A 1 oz. bullet fired horizontally with a velocity 1000 feet 
per second into a 1 lb. block of wood resting on a smooth table 
penetrates 2 inches and remains embedded. With what velocity 
does the block move off ? Would the bullet have penetrated more 
or less if the block had been fixed? 

7. An 1800 lb. shot moving with a velocity of 2000 feet per 
second impinges on a plate weighing 10 tons, passes through it 
and goes on with a velocity of 400 feet per second. If the plate 
is free to move find its velocity. 

8. Two inelastic bodies moving in the same direction with 
velocities of 10 and 8 feet per second impinge. If they weigh 

4 and 5 lbs. respectively, what is their common velocity after impact ? 
What would have happened if they had been moving in opposite 
directions ? 



CHAPTER IX 

STRESS AND STRAIN 

Strain may be defined as the change in shape or form of a 
body caused by the application of external forces. 

Stress may be defined as the force between the molecules of 
a body brought into play by the strain. 

An elastic body is one in which for a given strain there is 
always induced a definite stress, the stress and strain being 
independent of the duration of the external force causing them, 
and disappearing when such force is removed. A body in which 
the strain does not disappear when the force is removed is said 
to have a permanent set and such body is called a plastic body. 

When an elastic body is in equilibrium the resultant of all 
the stresses over any given section of the body must neutralise 
all the external forces acting over that section. When the 
external forces are applied, the body becomes in a state of strain, 
and such strain increases until the stresses induced by it are 
sufficient to neutralise the external forces. 

For a substance to be useful as a material of construction, 
it must be elastic within the limits of the strain to which it will 
be subjected. Most solid materials are elastic to some extent, 
and after a certain strain is exceeded they become plastic. 

Hooke's Law — enunciated by Hooke in 1676 — states that in 
an elastic body the strain is proportional to the stress. Thus, 
according to this law, if it take a certain weight to stretch a rod 
a given amount, it will take twice that weight to stretch the rod 
twice that amount; if a certain weight is required to make a 
beam deflect to a given extent, it will take twice that weight to 
deflect the beam to twice that extent. 

Kinds of Strain and Stress. Strains may be divided into 
three kinds, viz. (1) an extension; (2) a compression; (3) a slide. 



140 



STRESS AND STRAIN 



[CH. 



Corresponding to these strains we have (1) tensile stress; (2) co7n- 
pressive stress; (3) shear stress. 

A body that is subjected to only one of these, is said to be 
in a state of simple strain, while if it is subjected to more than one, 
it is said to be in a state of complex strain. 

Examples of simple strains are to be found in the cases of 
a tie bar; a column with a central load; a rivet, Fig. 75. The 
best example of a body imder complex strain is that of a beam 
in which, as we shall show later, there exist all the kinds of 
strain. 



Exicr 



■^ 



I 



ompres^ian 



\ "Iran&i^erBe, Sfnoin 



I 




RiveT under shear sTrxitn 
Kg. 75. Kinds of Strain. 

Intensity of stress. Imagine a small area a situated at a 
point X in the cross section of a body under strain, then if S is 
the resultant of all the molecular forces across the small area, 

— is called the intensity of stress at the point X. In the case of 

bodies under complex strain, the intensity of stress wiU be 
different at different points of the cross section, while in a body 
subjected to a simple strain, the stress will be the same at each 
point of the cross section, so that in this case if A is the area of 
the whole cross section and P is the whole force acting over the 

p 
oross section, the intensity of stress will be equal to -j . In future, 



IX] STRESS AND STRAIN 141 

unless it is stated to the contrary, we shall use the word *' stress " 
to mean the " intensity of stress." 

Unital strain. The unital strain is the strain per unit length 
of the material. In the case of extension and compression, the 
total strain is proportional to the original length of the body. 
Thus, a rod 2 ft. long will stretch twice as much as a rod 1 ft. 
long for the same load. In Fig. 75 if Z is the unstrained length 
of the rods under tension and compression and x the extension 

or compression, the unital strain is y . 

In the case of slide strain, the angle of the unit cube (Fig. 75) 
under consideration but not the length of the body is altered, and 
this angle j8 is the measure of the unital strain. If the angle is 
small, as it always will be in practice with materials of construction, 

X 

then it will be nearly equal to j , where x and I are the quantities 
shown in the figure. 

Stress-strain Diagrams. If a material be tested in tension 
or compression, and the strain at each stress be measured, and 
such strains be plotted on a diagram against the stresses, a 
diagram called the stress-strain diagram is obtained. If a material 
obeys Hooke's Law, this diagram will be a straight line. For 
most metals, the stress-strain diagram will be a straight line imtil 
a certain point is reached, called the elastic limits after which the 
strain increases more quickly than the stress, until a point called 
the yield point is reached, where there is a sudden comparatively 
large increase in strain. After the yield point is reached, the 
metal becomes in a plastic state and the strains go on increasing 
rapidly until fracture occurs. 

Fig. 76 shows the stress-strain diagram for a tension specimen 
of mild steel, such as is suitable for structural work. 

The portion AB of the diagram is a straight line, and represents 
the period over which the material obeys Hooke's Law. At the 
point C, the yield point is reached, and the strain then increases 
to such an extent that the first portion of the diagram is re-drawn 
to a considerably smaller scale, as shown on the left in the 
figure. The strain then increases in the form shown until the 
point D is reached, the curve between C and D being approxi- 
mately parabolic in shape. When the point D is reached, the 
maximum stress has been reached, and the specimen begins to 



142 



STRESS AND STRAIN 



[CH. 



pull out and thin down at one section, and if the stress is 
sustained, fracture will then occur. The portion DE, shown 
dotted, represents increase of strain with apparent diminution of 
stress. This diminution is only apparent because the area of the 
specimen beyond the point rapidly gets smaller, so that the load 
may be decreased and still keep the stress the same. In practice 
it is very difficult to diminish the load so as to keep pace with 
the decrease in area, so that this last portion of the curve is very 
seldom accurate, and has, moreover, little practical importance. 



o, 

CO 



m 

CI 



4^ 

r 



V) 



















^ 


.^ 


1 


"■X 


Mo xirftun 
\ 1 


S: 


■c> 


> 




















y 


y 






i 






'^fr 


.en 


're 


F 


vnT 












/ 


/ 










b 




75/ 


A 
'4 






















/ 


Tc 


K 


E, 


re. 


JS/ 


in 








1 




















/ 










n 






/I 


r 






































R 


7^ 


eld 


H 


jnf 
















, 


c 
















J 


















S 


• 


^ 


-y 


eJd 


\fi\inT 


■■"" 




""~ 




















y 


/ 


s». 


EL 


rsT/ 


,1 


,im 


t 








. 




, 












/ 


< 




































/ 


/ 




s 
































/ 


^ 


• X • 
1 


- -4 


-J 




























^ 


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J 

■0005 
•1 1 






'08lO 
•2 1 






•001 5 
'3 






■00 


20 



A Strain fior Inch LencfTh 



Frocture-d ForT'on 

Fig. 76. Stress -strain Diagram for Mild Steel in tension. 

The specimen draws down at the point of fracture in the 
manner shown in the diagram. Before the test, it is customary 
to make centre-punch marks at equal distances apart along the 
length of the specimen. The distance apart of these points after 
the fracture of the specimen indicates the distribution of the 
elongation at different points along the length. Four such 
marks, a,b,c, d, are shown in the figure. The greatest extension 
occurs at the point of fracture, so that with a specimen of short 
length, the percentage total extension will be greater than with 
a longer specimen. 



IX] 



STRESS AND STRAIN 



143 



The stress-strain diagrams in compression and shear for mild 
steel are very similar to that for tension. In compression it is 
difficult to get the whole diagram, because failure occurs by 
buckling, except with very short lengths, where it is very difficult 
to measure the strains, and in shear the test has to be made by 
torsion, because it is almost impossible to eliminate the bending 
effect. Now, in torsion, the shear stress is not uniform, so that 
the metal at the exterior of the round bar reaches its yield point 
before the material in the centre, and this has the effect of raising 
the apparent yield point. The same occurs in testing for com- 
pression or tension by means of beams. 

The importance of the elastic limit has been overlooked to 
a great extent by designers of structures and machines; but 
inasmuch as the theory, on which most of the formulae for 
obtaining the strength of beams are based, assumes that the 
stress is proportional to the strain, it must be remembered that 
our calculations are true only so long as Hooke's Law is true, 
so that the elastic limit of the material is a very important 
quantity. 

Stress-strain Diagrams for Cast Iron. The strength of cast 
iron varies largely with the composition, and the strength in tension 



Stf jin her a nt 

COMPRESSION 



•1 ^ 



y 




OAST IRON 



10 



5 /> 



Un 



TENSION' 



A. 



10 ^ 

CO 



15 
20 



1 .2 



Fig. 77. Stress-strain Diagrams for Cast Iron. 

is considerably less than that in compression. Fig. 77 shows the 
stress-strain diagrams for both tension and compression. It will 



144 



STRESS AND STRAIN 



[CJH. 



be seen that in tension the strain is never really proportional to 
the stress, while in compression the stress and strain are approxi- 
mately proportional up to a stress of about 8 tons per square 
inch. In the figure the compression curve is not completed, 
owing to buckling setting in. It is on account of the fact that 
the strain is not proportional to the stress that there is a consider- 
able difference between the actual and calculated strengths of 
cast iron beams. 

Other Materials. Timber. — There are several difficulties 
att-endant upon the accurate testing of timber, owing to the 
effect of dampness and to lack of homogeneity in the material. 
It may be taken that the stress-strain diagrams are approximately 
straight for a portion of their length, but then curve off in a 
similar manner to the compression curve for cast iron. 



2000 




•S 50 C 



yCOI^rCRETL IN COMPRESSION 



•0002 -0004 -0006 -0003 '00 10 ^12 
Shfin f>^r In. 

Fig. 78. Stress-strain Diagram for Concrete in compression. 

Cement and Concrete. — The stress-strain diagram for cement 
and concrete in compression is never exactly straight, so that 
there is no elastic limit, the exact curve depending on the 
composition and on the time after setting. 

The curve shown in Fig. 78 is almost exactly a parabola. 
This curve is for a 1 — 3 — 6 concrete, 90 days old, which was 
tested by Mr R. H. Slocom of the University of Illinois. Some 
authorities assume that the curve is a parabola but in practice 
it is seldom that the curve comes so near to a parabola as the 
above. The stress-strain curve is, however, nearly always of a 
similar shape, the strains increasing more quickly than the 
stresses. It is extremely important to remember that with 
cement and concrete the relations between stress and strain vary 
largely with the quality and proportions of ingredients, and 



IX] STRESS AND STRAIN 145 

cannot be taken as almost constant as in the case of steel. In 
tension a somewhat similar curve is obtained, but as cement and 
concrete are practically never used in tension, much less work 
has been done on its tensile strength. 

The Elastic Constants or Moduli, If a material is truly 

elastic, i.e. if the strain is proportional to the stress, then it 

follows that the intensity of stress is always a certain number 

, ,. ,- ..IX. ^1- X ^t. J.' intensity of stress 

of times the unital stram, or that the ratio rr^^ — i — ; 

unital strain 

is constant. This stress-strain ratio is called a modulus. That 

for tension and compression is generally known as Young's 

modulus, and is given the letter E; that for shear is called the 

shear or rigidity modulus (O). There is an additional modulus 

called the bulk or volume modulus {K) which represents the ratio 

between the unital change in volume and the intensity of pressure 

or tension on a cube of material subjected to pressure or tension 

on all faces. 

Young's modulus is the one which we shall be most concerned 

with in engineering design. Suppose a tension member (a 

tie as it is called) or a compression member (a strut) of length 

I and cross sectional area A is subjected to a pull or thrust P, 

and that the extension or compression is x, Fig. 75. Then the 

P X 

intensity of stress is -j , and the unital strain is ^ , 

P X PI 

:. Young's modulus = ^ = -7^1 — ~a~' 

The value of Young's modulus can be found from the stress- 
strain diagram. Thus in that for mild steel, Fig. 76, 

X 

stress 
Now in the relation E = -7 — ~-, if the strain is equal to 1, 

i.e. if the bar is pulled to twice its original length, we have that 
E = stress, and this accounts for the definition of Young's 
modulus that some writers have given, viz. " Young's modulus 
is the stress that is necessary to pull a body to twice its original 
length." Some students find this definition more clear than the 
one previously given, but it must be remembered that no material 

A. II. 10 



y 



146 STRESS AND STRAIN [ch. 

of construction will pull out to twice its original length without 
fracture. 

Numerical Example. A mild steel tie-bar^ 12 ins. long and 
o/ 1 J ins. diameter, is subjected to a pull of IS tons. Jf the eriension 
is '0094 in., find Young's modulus. 

Area of section of IJ ins. diam. = 1'767 sq. ins. ; 

18 
/. Stress per sq. in. = , ^^ = 10-19 tons per sq. in. 

Unital strain = -^ = -000783 ; 

.'. Young's modulus = ^ ^ = 13,000 tons per sq. in. 

Young's modulus for Concrete and similar Substances. If 
Young's modulus is a constant, it can be found for strains and 
stresses below the elastic limit only, and, strictly speaking, there 
is no modulus for substances such as concrete, where the strain 
is not proportional to the stress. From Fig. 78 it is clear that 
since the strain increases more quickly than the stress in concrete, 

the value of the ratio —r — ;— will be greater for small stresses than 

stram ° 

for large stresses, and so, before the value of this ratio is of any 

use to us, we must know the value of the stress at which the 

ratio is calculated. One can hardly lay too great stress on the 

importance of having exact ideas on the principles which form 

the foundations on which the theory of structures is built, and 

with concrete it is practically useless to speak of the compressive 

strength and Young's modulus unless the composition of the 

concrete and the stress at which the modulus is calculated are 

known. 

Poisson's Ratio — Transverse Strain. When a body is 

extended or compressed, there is a transverse strain tending to 

prevent change of volume of the body. The amount of transverse 

strain bears a certain ratio to the longitudinal strain. 

^, . . . transverse strain . . -i . ^ t 

This ratio = , ^r— .p — = — r — r- = rj vanes from 4 to f for 

longitudmal stram ' 

most materials, and is called Poisson^s ratio. 

According to one school of elasticians, the value of this ratio 

7) should be J , but experimental evidence does not quite support 

this view, although it is very nearly true for some materials. 

The ratio is very difficult to measure directly. 



IX] 



STRESS AND STRAIN 



147 



Experiment upon the tensile strength of wire. Although it requires 
a heavy testing machine to make some experiments upon the strength of 
materials, the following comparatively simple apparatus wiU enable a good 
deal to be learnt concerning the tensile strength of wire. 



/ft) /ft\^ flh 

R oof Member \^^^^^^^p^ 





I 



A wire about 8 feet long is suspended from the ceiling or other convenient 
point as shown in Fig. 79, and a scale pan is hung on the end. Suspended 
from the same point is a strain-measuring device or extensometer constructed 

10—2 



148 



STRESS AND STRAIN 



[CH. 



conveniently as shown in Fig. 80. A white celluloid vernier is carried by 
a slider which is clipped to the wire at a fixed length I from the point of 
suspension, this length being called the gauge length. By means of this vernier, 
the length I for dififerent loads upon the wire can be calculated. The following 
records of an experiment made with this apparatus illustrate its use. Weights 
are added gently to the scale pan and the scale and vernier are read after 
each addition. The load is then the weight of the scale pan + the total weight 
added as will be followed from the third column: 

Material — Soft Iron. Gauge length — 84*. 
Diam. (mean of six measurements) 'OSSi'. 
Weight of scale pan 0-6 lb. 



Observed 


Calculated 




Added 


Scale 


Load 


Extension 


wts. 


reading 


lbs. 


ins. 


NOTES 


3 


•65 


3-5 


•013 


3 


•66 


6-5 


•023 




2 


•67 


8-5 


•033 


Extension is scale reading less 


2 


•675 


10-5 


•038 


•637, the correct zero as 


2 


•68 


12-5 


•043 


determined from the pre- 


2 


.685 


14-6 


.048 


liminary plotting (Fig. 81.) 


2 


•69 


16-5 


•053 




2 


•695 


18-5 


•068 




2 


•70 


20-5 


•063 




2 


•71 


22-6 


•073 




2 


•72 


24-5 


•083 




2 


•725 


26-5 


•088 




2 


•73 


28-5 


•093 




2 


•74 


30-5 


•103 




2 


•84 


32-5 


•113 




2 


1-40 


34-5 


•763 




2 


2-20 


36-5 


1-563 




2 


2-65 


38-5 


2013 




2 


3-23 


40-5 


2593 




2 


3-9 


42-5 


3-263 




2 


4-65 


44-5 


4013 




2 


56 


46'G 


4-963 




2 


6-§ 


48-5 


6-163 




2 


8^4^ 


50-5 


7-813 




2 


11-2 


52-5 


10-563 




1 


13-8 


53-5 


13-163 


Broke. 



K] 



STRESS AND STRAIN 



149 



/ 



o 

-J ? 



] I 







_, V y 




Sectigh on A.B. 



Fig. 80. 




























*" 


-^ 1 


_ 


7) 


:5 


~ 




-5 


^ = 


: 


.^i 


— 
_ 


~ 


ii^ 


<0 z 


— 


^t 


- 












Clip 
' (Sttcl) 




^Cllp 
Detail of Face 

of Slide r 
Scale full Siz0 



Results. 
Initial Area = '785 x (-0364) 2 = -00104 sq. in. 
Final Area = 785 x ( 0334)2 = -000876 sq. in. 

0925^ .^101= 24,400.000 lbs, per sq.m. 



^ = — X - = 
a A 



63-5 



•00104 



Breaking Stress = 77^7^= 51,400 lbs. per sq. in. 

= 22-9 tons per sq. in. 



Stress at Elastic limit = 



29 



QQIQ4^= 27,900 lbs. per sq. in. 

= 12-4 tons per sq. in. 

Elastic limit Stress 29 
Breaking Stress 63 5 ■— 

Extension = ^ = 15'7 percent* 



150 



STRESS AND STRAIN 



[OH. 



From the observed results the preliminary stress-strain diagram shown 
in Fig. 81 is first plotted. This enables us to find the zero reading on the 




•7 -75 

Scale- //VC//f5.> 

Fig. 81 



60 




























50 


















, — 




-^ 




. ■ 










,^ 




"^ 














40 




i 


\ 
X 


^ 


r 




















y 


/ 






















30 


» 
















^ 




-"^ 






' 












y 


K 




f.I.- 


29Z.i 


r5. 




20 


» 










y 


y^ 


1 
1 

1 




















y 


/■ 




1 

0^1 












10 






V 


y 

^ 








Si 














.^ 


/o 








•0 


1 

1 

925 
















^ 




•0 


* 






! 


E> 


^LARC 


LD L 


XT.S 


CAIE 



& 8 

ExrENs/ONS Ins. 



10 



12 13 



Fig. 82. 
scale from which we are able to calculate the extensions given in the fourth 
column and thus to plot the complete stress-strain diagram shown in Fig. 82. 



IX] STRESS AND STRAIN 151 

This is strictly a diagram of loads plotted against extensions and is not one of 
intensity of stress plotted against unital strains. The real stress-strain diagram 
will have the same shape but a different scale so long as the area remains 
constant. In practice the breaking stress is always calculated by dividing the 
breaking load by the original area. 



Factor of Safety. We will use the term ** factor of safety " to 

J ^ XI J.' Breakinor stress of material mi • • ^i. 

denote the ratio == — r-- —. 1—- — 5 — = — • This is often 

Working stress used m design 

, , ,, Load required to cause failure ,.,, , 

taken the same as K= , 7-3 . The two 

Load carried 

results are not the same because the theories that we use in 

design do not allow for all the possible contingencies. When 

using the second definition it is usual to specify a factor of safety 

of 4 for structures not subjected to impulsive loads and vibrations. 

The term "factor of safety" is used very loosely in practice 
and it is very desirable to use it in some definite sense as the 
first one given above. 

The great difficulty of getting a satisfactory definition of the 
factor of safety resides in the fact that we can find the load to 
require failure of a structure only by means of a test to destruction 
which is too costly and defeats its own end. Li machine parts the 
same difficulty does not arise because the cost of a part to be 
tested will be small ; the difficulty there is that the part must be 
tested under the actual conditions in which it is used in practice. 
For instance it is no use testing motor car axles to get an accurate 
value of the factor of safety by making a tension test on a piece 
of the material. Since the elastic limit in steel is the poiat which 
determines the real safety of a structure it would be more satis- 
factory to define the factor of safety in steel and other metals 
possessing an elastic limit as the ratio 

Maximum calculated stress in material 
Elastic Limit stress in material 

Working Stresses. In the absence of other regulations it 
is usual to take the working stresses as J of the breaking stresses 
(see table on p. 157) but for cast iron J is often taken. For mild 
steel, wrought iron and cast iron the following values may be 
safely taken where the body is not subject to shocks and vibra- 
tions. 



152 



STRESS AND STRAIN 



[CH. 





Working Stresses in tons per sq. in. 


Tension 
Si 


Compression 

/e 


Shear 


Mild steel 
Wrought iron 
Cast iron 


7 
6 
1 


6 
4 
6 


6 
4 
1 



Work done in Straining; Resilience. If we look at a stress- 
strain diagram, we shall see that the strains are, to a reduced 
scale, the total distance moved by the end of the specimen 
because extension = distance moved = unital strain x original 
length. Also in simple stressing we have 

.*. F = Af, but F is equivalent to the force that we have 
previously spoken of as the effort, so that if the area keeps con- 
stant — as it does for all practical purposes within the elastic 
limit — the stress is a measure of the effort. 







^/, 


p t 


c 




^.'^^/yvvO 




^ 


^^ 


^^^ 


F= 


t 




<i JC 


> 


M 



Extension 

Fig. 83. Resilience. 

The stress-strain diagram therefore is a special kind of 
diagram of effort plotted against distance and we have previously 
called such a diagram the effort curve (p. 43). 

We have proved on p. 43 that the work done is equal to the 
area below the effort curve ; therefore the area below the load- 
extension diagram must give the work done in straining the bar. 
Now the work done in straining a material per unit volume of the 
material is called the Resilience 

Referring to Fig. 83, let I be the original length of the bar. 



IX] STRESS AND STRAIN 

The work done in producing a stress / 

= Area of A POM 
= IPM . MO 
= \F .X 
= \f.A X. 

L 



153 



,, X ..IX- stress 

Now J == unital strain = — ^p- 



X = 



E' 



Work done in producing a stress / 











Al = 


volume of the bar = F, 




Work ^ .,. P 

:. ^^^-i = Resilience = ^-^ . 

Volume %E 



Stresses and Strains due to Sudden or Dynamic Loading. 
If a load is applied suddenly to a machine or structure, vibration 
will ensue, and the strain — and thus the stress — will reach twice 
the value which would occur if the load were gradually applied. 




This will be made clear from considering a diagram, Fig. 84, 
where the force is plotted against the strain. We have seen that, 
with gradual loading of an elastic body, the curve representing 
the relation between the strain and the load in direct stress is 
represented by a straight line AD, the area below the line giving 
the work done up to a given point. Now let AG represent a 
force P; then when the strain gets to the point B, the work 



154 



STRESS AND STRAIN 



[CH. 



don© by the force will be equal to the area of the rectangle 
ABEGy whereas the work done in straining the material is only 
equal to the area of the triangle ABE, so that there is an amount 
of work equal to the area of the triangle AEG still available for 
causing increased strain. The strain therefore increases until the 
area of the triangle EFD is equal to that of the triangle AEG. 
It is clear that AG = 2^JB, or that the strain — and thus the stress 
— is twice that in the case of gradual loading. 

This is a most important point and shows us that we should 
make allowance in engineering calculations for the nature of the 
loading, whether gradual or sudden. In the latter case therefore 
we ought to allow a greater factor of safety. 

Gradual loads are usually called "static loads'* or "dead 
loads,'* and sudden loads are called "dynamic loads" or **live 
loads." 

It is a good rough rule to take a live load as equivalent to 
a dead load of twice its value. 

Strain and Stress due to Impact. Suppose a weight W falls 
from a height ^ on to a structure and let the deformation or 
strain in the direction of h \)g x, Fig. 85. Then the work done 



(W) 



'111 







Stra/n 
Fig. 85. 

by the weight is equal to W {h + x). Now this work is absorbed 
in straining the structure. Consider first the case in which the 
resulting strain is within the elastic limit. The work done in 
such case is equal to the volume multiplied by the resilience. 
We have shown that in tension or compression the resilience is 

equal to „ ,, and therefore in this case we get 

Volume X P VP 



W {h+x) = 



2E 



2E 



IX] STRESS AND STRAIN 166 

Then if ar is negligible compared with h we have 



/■■ 



2EWh 
If the weight strikes with a velocity v, 

h "' 

2gr' 
. /2E . Wv^ /EW 

Strain beyond Elastic Limit. If the strain is beyond the 
elastic limit, it follows, from the reasoning given on p. 153, that 
the work done per unit volume in straining is equal to the area 
below the stress-strain curve. If this area is jR, Fig. 85, then we 

have R = Wh or -»— . 

From this the stress can be found. 

Numerical Example. A bar of J inch diameter stretches 
J inch under a steady load of 1 ton. What stress would be produced 
in the bar by a weight of 150 lbs. which falls through 3 inches befwe 
commencing to stretch the bar — the bar being initially unstressed 
and the value of E taken 05 30 x 10^ lbs. per square inch ? 

Area of bar V diam. = -196 sq. in. 

.*. Stress under load of one ton = -^7^7^ tons per sq. in. 

•196 

2240 „ 
= typ^ lbs. per sq. m. 

^, . Stress 2240 

.*. Stram = 



E 196 X 30 X 10«* 

Now Y = Strain x Original length, 

^ . . , , ., I -196 X 30 X 106 

.*. Origmal length = ^i^ . = t^ktk — 7. — 

^ ^ Stram 2240 x 8 

.*. Volume = Length x Area of section 

= -196 X -196 X 30 x 10« 

8 X 2240 ' 
«= 64' 31 cub. ins. 



166 STRESS AND STRAIN [ch. 

Work done by 150 lbs. in falling 3 inches = 3 x 150 = 460 in.-lbs. 

.^•31x/^^ 

2E -*°"» 

/ 900E 
^"W 64-31 



-y 



900 X 30 X 10« 



64-31 
= 20,490 lbs, per sq. in. Ans 

Temperature Stresses. Suppose a bar of length I is heated 
t° F. and a is coefficient of expansion. Then, unless prevented, 
the length of the bar will become I (1 -\- at)y i.e. the increase in 
length will be atl. 

If the bar is rigidly fixed so that this expansion cannot take 
place, then there wiU be in the bar a strain equal to atl, and the 

unital strain will he -j- = at. 

This strain will produce a compressive stress of at x E, where 
E is Young's modulus. 

Now for mild steel a = -00000657 per degree Fahrenheit, and 
E = 13,000 tons per square inch. 

/. The stress per ° F. = -00000657 x 13,000 

= -0854 ton per square inch. 

Taking a range of temperature of 120° F., the stress due to 
temperature = 120 x -0854 = 10*25 tons per square inch. This 
is more than the safe stress for mild steel, so that the importance 
of designing structures so that the expansion may take place 
becomes quite evident. 

Struts, Columns and Pillars. When bars are in compression 
they are called struts, columns or pillars; under test such bars 
always fail by buckling as indicated in Fig. 86, buckling being 
a kind of bending. Full consideration of this question is rather 
difficult and beyond our present scope. We will just point out 
the following facts : 

(1) The strength of a strut depends upon its length and shape 
of cross section as well as upon its cross sectional area. 



IX] 



STRESS AND STRAIN 



157 



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158 



STRESS AND STRAIN 



[CH. 



(2) In choosing a column section we should try to get it 
about as broad one way as the other and as much of the material 
should be as far as possible from the centre. For the latter 
reason, shapes (6) and {d) are better than (a) and (c) but all are 
fairly suitable. 





Fig 86. 



SUMMARY OF CHAPTER IX. 

Strain is the change in shape or form of a body caused by the 
appHcation of external forces. 

Stress is the force between the molecules of a body brought into 
play by the strain. 

Hooks' s Law states that in an elastic body strain is proportional 
to the stress. 

There are three kinds of stress : tension, compression and shear. 

Intensity of stress is the total stress on a small area divided 
by the area. 

Unital strain is the strain per unit length of the material. 

The elastic limit of a material is the stress at which the strain 
ceases to be proportional to the stress. 



IX] STRESS AND STRAIN 159 

The yield point is the stress at which the strain increases suddenly 
witiiout increase of stress. 

Y , J , y,_ Ten sile or comp ressive intensity of stress 

* Tensile or compressive unital strain 

Ax' 

-, . £ t A. Breaking stress 

Factor of safety = 



Poissori's ratio = 



Working stress used in design * 
Transverse strain 



Longitudinal strain * 



Resilience is the work done in straining a material per unit 
volume. 

Gradually applied loads are called dead loads ; suddenly applied 
ones are called live loads. 

A live load causes twice the stress caused by a dead load of the 
same amount. 



EXERCISES. IX. 

1. A tie -rod in a roof whose length is 142 ft. stretches 1 inch 
when bearing its proper stress. What strain is it subjected to? 

2. How much will a tie-rod 100 ft. long stretch when subjected 
to -001 of strain? 

3. A cast iron pillar 18 ft. high shrinks to 17*99 ft. when loaded. 
What is the strain? 

4. A tie-rod 100 ft. long has a sectional area of 2 sq. ins., it 
bears a tension of 32,000 lbs. by which it is stretched f . Find the 
intensity of the stress, the strain, and the modulus of elasticity. 

5. How much will a steel rod 50 ft. long and \ sq. in. sectional 
area be stretched by a weight of one ton, the modulus of elasticity 
being 35,000,000 lbs. per sq. in. ? 

6. Find the work done in stretching each of the rods in Questions 
4 and 5. 

7. The diameter of the piston of an engine is 12', the diameter 
of the piston-rod being 2Y' Find the stress in the piston-rod 
when the maximum steam pressure is 120 lbs. per sq. in. 

8. Find the proper diameter for a wrought iron rod to sustain 
a direct pull of 1 3 tons, the greatest stress allowable being 9000 lbs. 
per sq. in. 



160 STRESS AND STRAIN [ch. ix 

9. The diameter of a steel rod is 2 ins., find the greatest weight 
it could support so as not to stress it to more than 10,000 lbs. per 
sq. in. 

10. What stress in lbs. per sq. in. will stretch an iron bar 
whose length is 12 ft. by I in. ? The modulus of elasticity of iron 
being 28,000,000 lbs. per sq. in. 

11. Define stress, strain and modulus of elasticity. The 
cross sectional area of a piece of wire is -02 sq. in. and its length 
is 20 ft. When loaded with 160 lbs. it stretches -08 in. Find the 
modulus of elasticity of the wire. 

12. Find the work done in stretching the bar in Question 10. 

13. An iron rod is suspended by one end. Draw a curve 
showing the stress at any section, and find the length of a rod which 
can just carry its own weight, allowing a working stress of 9000 lbs. 
per sq. in. and weight of material 480 lbs. per cubic ft. 

14. The elastic limit of a bar was found to be 20,000 lbs. per 
sq. in. and the strain at this point was -0006. What was the resilience 
of the material? 

15. A load of 560 lbs. falls through | in. on to a stop at the 
lower end of a vertical bar 10 ft. long and 1 sq. in. in section. If 
E = 13,000 tons per sq. in. find the stress produced in the bar. 

16. A tie -rod in a roof structure has to stand a total pull of 
40 tons. If the breaking stress in the material is 30 tons per sq. in. 
and a factor of safety of 6 is required, find a suitable diameter 
for the bar. 

17. What is Poisson's ratio ? If a steel bar 3 inches in diameter 
and 6 inches long is subjected to an axial pull of 70 tons, find the 
longitudinal and transverse strains if E — SO x 10* lbs. per sq. 
in. and »? = 4. 

18. A cylinder cover 10 inches in diameter is attached by 12 
studs f inch diameter at the bottom of the thread (-3 square inch 
in area). Find the force per square inch in these, when the steam 
pressure is 100 lbs. per sq. in. by gauge. 



CHAPTER X 



RIVETED JOINTS; THIN CYLINDERS 

Forms of Rivet Heads. The most common forms of rivet 
heads and their usual proportions are shown in Fig. 87. 



OTJP OB SKAP HEAD 




CONICAL HEAD 




COUNTKBSUNK HEAD 



Fig. 87. Forms of Rivet Heads. 

For structural work the snap-headed rivets are most usual, 
but countersunk rivets are used where necessary to prevent pro- 
jections from the surface of the plate. Snap -heads take a length 
of rivet equal to about 1 J times the diameter. 

It is usual in practice to adopt a diameter of rivet when 
cold equal to one-sixteenth of an inch less than the diameter of 
the hole, but in all calculations the diameter of the rivet is taken 
as being equal to that of the hole. 

Diameter of Rivets. According to Unwin's formula, the 
diameter of the rivet is 1*2 Jty where t is the thickness of the 
thinnest plate, but for structural work this rule is very seldom 
adopted. In practice a f or J" rivet is used wherever possible, 
and it is best not to use any formula to obtain the diameter 
in terms of the thickness of the plate. Some authorities use 

A.M. 11 



162 



RIVETED JOINTS; THIN CYLINDERS 



[CH. 



a diameter of f for a f pJate, f for a J' plate, and 1" for 
a f plate. It is difficult to get rivets of larger diameter than 
1 in. driven by hand. 




"^ (xyy 



W/Z/A'. 




■9^^ 



LAP JOINT. 




^— 



BUTT JOINT. 



-^^ 




^ 




^^- 



SINGLE COVER JOINT. 
Fig. 88, 



X] 



RIVETED JOINTS; THIN CYLINDERS 



163 



It is a mistake to adhere too rigidly to Unwin's formula ; the 
best diameter wiQ be that which will give equal shearing and 
bearing strengths (see p. 164 for explanation of these terms) and 
will therefore depend upon whether the rivets are in double or 
single shear. Basing our ideas upon this we should get the 
following formulae: 

d = 2'5t for single shear 
= 1*25^ for double shear. 

For single shear this would make d so large for thick plates 
that there would be practical difficulty in heading the rivets. 

Forms of Joints, (a) Lap Joints and Butt Joints. In 
the lap joint the plates overlap as shown in Fig. 88. This form 
of joint has the disadvantage that the line of pull is such as to 
cause bending stresses, tending to distort the joint as shown. 

In the butt joint the edges of the plate come flush, and cover 
plates are placed on each side as shown, the thickness of the 
cover plates being each five-eighths that of the main plates. In 
this form of joint the pull is central, so that there are no bending 
stresses. 

In the single cover joint, which is a cross between the lap joint 
and the butt joint, there are bending stresses developed, tending 
to distort the joint as shown. 

It is clear from the above that the butt joint should be adopted 
wherever possible. 

(6) Chain Riveting and Zig-zag or Staggered Riveting. 









o 


o 


o 


o 


o 


o 


o 


o 


o 


o 


o 


o 










Fig. 89. Chain riveting. Fig. 90. Zig-zag riveting. 

The different rows of rivets in a joint may be arranged in chain 
form or zig-zag form, as shown in Figs. 89, 90. As we shall see 

11—2 



164 RIVETED JOINTS; THIN CYLINDERS [ch. 

later, the zig-zag form is more economical, and should be used 
whenever possible. 

Methods in which a Riveted Joint may Fail. A riveted joint 
may fail in any of the following ways : 

(1) By tearing of the plate. 

(2) By shearing of the rivets. 

(3) By crushing of the rivets. 

(4) By bursting through the edge of the plate. 

(5) By shearing of the plate. 

Fig. 91 shows these methods of failure. 

(4) and (6) are allowed for by the following rule : The minimum 
distance between the centre of a rivet and the edge of the plate 
is l^dy where d is the diameter of the rivet. 

If this rule is adhered to the joint will always fail first in one 
of the ways (1), (2), (3). 

The aim in designing a joint should be to make the force 
necessary to cause failure in the various ways equal. 

We will now consider the various ways of failure in detail, 
taking in each case a strip of plate equal to the pitch of the rivets. 

(1) Teabing of the Plate. In this case the width along 
which fracture will occur is {p ^ d), and as the thickness of the 
plate is t, the area of fracture = {p — d) t. 

Therefore it ft is the safe tensile stress in the material, the safe 
load which the joint can carry is equal to 

P=ft{p-d)t (1). 

(2) Shearing of the Rivets. 

In the case of single shear, the area sheared = ~^ , 

„ „ double „ „ „ = ~— . 

Therefore if fg is the safe shear stress on the rivet, the safe 
forces on the joint as regards shear are respectively 






(2). 



4 J 

♦ A Board of Trade rule makes this TTS instead of 2, this being based 
upon experiments. The rule is generally used in boilers but not so much in 
bridge work. 



X] 



RIVETED JOINTS; THIN CYLINDERS 



165 



(3) Crushing or Bearing of Rivets. In this case the 
crushing or bearing area is taken as the diameter of rivet multi- 
plied by the thickness of the plate, i.e. d x t. Therefore, if /, 




I 



/ 






! 



*P 



I 



a 



( 

/I 

( 



S/NGt£ SHCAR 



© 



/ 




) 



DOUBLE 9HCAR, 



t-. 



-®- 



- P 



© 



iP 



-JU- 




iP 



^ 



— p 




Fig. 91. Methods of failure of a Biveted Joinfi^ 



166 



RIVETED JOINTS; THIN CYLINDERS [ch. 



is the safe bearing stress on the rivet, the safe force on the joint 
as regards bearing is equal to 

The values of ft and /, may be taken as given in Chapter IX. 

For/e, 10 tons per square inch may be taken for mild steel, 

and 8 tons per square inch for wrought iron. These figures are 

higher than for ordinary compression, and are obtained from the 

results of experiments. 

For structural work the strength of the joint as regards bearing 
will often be less than as regards shear, because the plates are 
often thin compared with the diameter of the rivet, but this does 
not so often occur in boilers. 

Efficiency of Joint. The efficiency of a joint is the ratio of 
the least strength of a joint to that of a solid plate, i.e. 
^on • _ _ Least strength of joint 

^ ~ Strength of solid plate * 
The aim in the design of riveted joints is to get the efficiency 
as high as possible. 

Numerical Examples. (1) A plate 10" wide and f thick 
is jointed by a single riveted lap joint with rivets of \\ ins. diameter. 
If the tensile breaking strength of the plate is 22 tons per sq. in. and 
the shear breaking strength of the rivets is 20 tons per sq. in., how 
will the joint fail and what is its efficiency ? 

Width across which the joint will tear is (10 - 3 x 1 J) ins. 
.*. The tearing area = (10 — 3 x IJ) f sq. ins. 
.'. Force required to tear 

- (10 - 3 X 1 J) I X 22 *= 109 tons approx. 



4i 



\y: 



- r 



4 



7~T 

10" 




r, — X — ^ ' 

\VbDiam. ' |34' 

Fig. 92. 

Each rivet is in single shear. 

.*. Force required to shear rivets 

•= 3 X ^ X (I) X 20 = 69-5 tons. 



X] RIVETED JOINTS; THIN CYLINDERS 167 

This is less than that for tearing. 

.*. The joint will fail by shearing of the rivets. 

TT-ffi • _ I^^^st st rength of joint 

^ ~ Strength of original plate 

59-5 59-5 



10 X I X 22 165 
= '361 = 361%. 

(2) Design a dmihle-riveted lap joint to connect two steel plates 
J in. thick with steel rivets. The tensile strength of the plates before 
drilling being 30 tons per sq. in. ; the shearing strength of the rivets 
24 tons per sq. in. ; and the compressive strength of the steel 43 tons 
per sq. in. Find the efficiency of the joint. 

For J in. plates Unwin's formtda would give 

d = 1'2 J-5 = '85 in., say J in. 

The joint is a double-riveted lap, therefore there will be two 
rivets in single shear in a width of plate equal to the pitch. 

.*. Strength against tearing per pitch 

= ft(p-d)t 

= S0ip-d)i^l5(p-d) . .(1). 
/. Strength against shearing per pitch 

27rd2 



= /,. 



4 

24.27r 



If these are equal, 

7^ 



15(p-g) 



4 
= 289 tons. 

28-9; 



•©' •■— = <2) 



28-9 7 
•'•^~ 15 "^8 

= 1-93 + -87 = 2-80, say 3 ins. 

The bearing stress for a force of 29-8 tons would be equal to 

28-9 _, 
1x^x2 ^ ^® P®^ ^^' ^^'' 

the bearing area of each rivet being I x J = '437 sq. in. 



168 



RIVETED JOINTS; THIN CYLINDERS 



[CH. 



This is less than the allowable value of 43 tons per sq. in., 
showing that a larger diameter of rivet might be used with greater 
economy, but J in. diameter is in most cases more suitable in 
practice. 

The efi&ciency of joint in this case is equal to 



28-9 



28-9 



30 X 3 X ^ 45 



= 64-2%. 



The joint is then as shown in Fig. 93. 







i 



IT 



Fig. 93. 



(3) Find the number of rivets necessary to conned the gusset 
plates J etc., at the base of a steel stanchion to the stanchion proper, 
the load carried being 150 tons. The diameter of the rivets is J in. 
and the thickness of the plate J in. 

The rivets are best designed in such cases to carry the whole 
load, so that if the stanchion itself does not bear on the base plate 
the rivets will distribute the load satisfactorily. 

The strength of each rivet in single shear 



= ~ . Q^ 5 = 3-01 tons. 



The strength of each rivet in bearing 

= ^ . ^ . 10 = 4-37 tons. 

150 
c*o Number of rivets necessary = ^t^t^ = 50 nearly. 



X] 



RIVETED JOINTS; THIN CYLINDERS 



169 



WOKKING SlRENaTH OF StBBL RiVETS. 



Diam. 

of 

Biyets 

in 

infl. 


Area 
in 

sq. ins. 


Strength 
in single 
shear at 
5 tons 
per sq. in. 


Bearing Strength at 10 tons per sq. in. 


Thickness in ins. of plate 


^ 


f 


^ 


i 


A 


f 


H 


f 


f 

f 
1 
1 

1 


1104 
•1963 
•3068 
•4418 
•6013 
•7854 


•56 
•98 
153 
2-21 
301 
3-93 


117 
1-56 
1-95 

2-34 

2-72 
312 


141 

1^87 
2-34 
281 
3-27 
3-75 


1^64 
218 
2-72 
3-27 
3-82 
4-37 


1-87 
2-50 
312 
3-75 
4-37 
500 


211 

2-81 
3-51 
4-21 
4-91 
5-62 


2-34 
312 

3-90 
4-69 
5-46 
6-25 


2-59 
3-43 
4-30 
516 
602 
6-87 


2-81 
3-75 

4-68 
5-63 
6-56 
7-50 



The Strength of Thin Cylinders and Pipes. Suppose that a 
thin pipe or cylinder such as is shown in Fig. 94 carries inside 
it a fluid such as air, steam or water under pressure. There are 
two principal ways in which it might fracture; firstly longi- 
tudinally as indicated at the bottom and secondly circum- 
ferentiaUy at a Hne such as XX. 

Longitudinal strength. We will first consider the stresses in 
a longitudinal section and will neglect the additional strength 
given by the two ends. Take for instance a length I of the pipe. 
The fluid under a pressure p lbs. per sq. in. acts radially all round 
the section as shown on the upper portion of diagram and the 
total pressure acting on the upper half of the length I will be 

Trdl 



equal to pressure x area 



X p, but it is only the vertical 



component of the pressures which tends to cause the fracture 
under consideration. In the case of the pressure oa, for instance, 
ab is the part which is effective ; we therefore get the result that 
the effective total pressure tending to burst the pipe is given by pdl, 
i.e. the force obtained by considering the pressure as acting upon 
the diameter. 

If /^ is the tensile stress at the section and A is the area of 
fracture we shall therefore have 

ff.A =pdl; 



170 RIVETED JOINTS; THIN CYLINDERS [oh. 



a- 



7 0^--- 


-\b \ 


1 A ^ ^ /^ /|^ /fv /Tv ^ 

k d 


''] 



i 



riiiuiuuiu!nii/iU!Jiuuwriii'UiiunmuurrwTtr. 



s 



d 



U— I —=A 



'm^niiinmninnvnririUUiinnnuuiiW'ni^^ 




Fig. U4. Thin Cylinders and Pipes. 



X] RIVETED JOINTS; THIN CYLINDERS 171 

but the area of fracture is a rectangle at each side of length I and 

thickness t, 

i.e. A = 2U ; 

.-. /, . 2lt = pdl. 

Circumferential Strength. The force tending to cause fracture 
in this case is the total bursting pressure upon the ends, 

I.e. p X area of pipe = ~ — -^ — . 

The area of fracture is a thin circular ring of diameter d and 
thickness t; this area is for all practical purposes equal to 
circumference of inside of pipe x thickness of pipe = Trdt ; 

.*. Stress X area of fracture = total bursting pressure, 

p X ird^ 



4 

p X trd^ 



i.e. fi X irdt = 

/, 

'± A TTU/fr 
/n/7 

(2). 



4 X Tcdt 
_pd 
~ Tt 

This is exactly one -half of the stress in the longitudinal 
section and shows that the longitudinal section is the weakest. 
In the design of boiler shells therefore the longitudinal seams 
are provided with stronger joints than the circumferential seams. 

By means of formula (1) we can find the thickness required 
for a pipe of given diameter to withstand a certain pressure and 
keep the stress within certain fixed limits. 

Numerical Example. How thick would you make a boiler 6 feet 
in diameter which has to withstand a pressure of 200 lbs. per sq. in., 
if the stress in the material must not exceed 12,000 lbs. per sq. in, ? 

From equation (1) 

f -^ 
Jt 2t ' 

ft = 12,000, p = 200, d = 12 inches; 

200 X 72 



12,000 = 



2t 



, 200 X 72 ^ . , 

^ = 2-^^2:000 = "^^^'^' 

say fin. 



172 RIVETED JOINTS; THIN CYLINDERS [ch. 



SUMMARY OF CHAPTER X. 

Riveted joints may be lap joints or butt joints. 
Shear strength of a rivet 

*= ^^~- for single shear, 

= 1-75/.^^ 



^ for double shear. 



Bearing strength of a rivet 

= fs • d »t, 

^^ . - • X J • • X Least strength of joint 

Efficiency of a riveted Jomt = g^^;^p^^^^^^j^^-p^. 

The circumferential strength of a thin cylinder is twice the 
longitudinal strength. 



For longitudinal strength 






EXERCISES. X. 

1. If the ultimate shearing strength of a steel plate is 20 tons 
per sq. inch, what force will be required to cause a 1 inch rivet in 
a single -riveted lap joint with a lap of 2 inches to shear through 
the plate which is f in. thick ? 

2. What diameter of rivet would you use for a ^ inch plate, 
and what pitch would you use for a lap joint? What would the 
efficiency of your joint be? 

3. In a butt joint with a single row of rivets the plates are ^ inch 
thick, the rivets are | inch diameter and IjJ inches apart; calculate 
the efficiency of the joint. 



xj 



RIVETED JOINTS; THIN CYLINDERS 



173 



4. Two rectangular tie-bars are united by two cover-plate^- 
as shown (see Fig. Xa). If /,=/<= 6 tons per sq. in., find the 
resistance to shearing of the rivets and to tearing of the plates. 

What should be the thickness of the cover-plates? What is 
the efficiency of the joints? 




Rivets l"diam. 



Fig. Xa. 

5. How many rivets would you use to connect a member of 
a roof -truss to the main body of the truss if the member carries 
20 tons and is ^ in. thick? The rivets are | in. diameter. Is bearing 
more important than shearing in this case? 

6. What load may be safely carried by a column which has 
40 rivets in single shear connecting the column to its base ? The 
rivets are f in. diameter. Take a safe shear stress of 5 tons per sq. in. 

7. For equal strengths in tension and shear calculate the pitch 
for a butt joint given the following data. Plates 1 inch thick; 
rivets 1 J ins. diameter ; two rows of rivets on each side of joint ; 
fa = 64,000 lbs. per sq. in. ; /« = 65,000 lbs. per sq. in. 

8. A cylindrical boiler 8 feet in diameter is to withstand a 
worldng pressure of 100 lbs. per square inch. Calculate to the 
nearest \ inch the thickness of the shell, allowing a stress of 10,000 lbs. 
per square inch, and neglecting the effect of the joint. 

9. Find the thickness of an iron boiler shell to withstand an 
internal pressure of 250 lbs. per sq. in. The diameter is 10 ft. and 
safe stress allowable on plates 3 tons per sq. in. 

10. What should be the thickness in the previous question 
if the joint ha^ an efficiency of 60 per cent., the shell being composed 
of two plates? 

11. Find the safe working pressure in lbs. per sq. in. for a boiler 
6ft. diameter, plates Y thick ; allowing a stress of 3500 lbs. per sq.inch. 

12. The diameter of a steam boiler is 4 ft. Find the thickness 
of plates necessary if the stress is not to exceed 6 tons per sq. in. 
when the internal pressure is 120 lbs. per sq. in. and the efficiency 
of the joint is 72 per cent. 



CHAPTER XI 

THE FORCES IN FRAMED STRUCTURES 

A THEORETICAL framed structure is built up of a number 
of straight bars, pin- jointed together at their ends. If the centre 
lines of the bars all lie in the same plane, the frame is called 
a plane frame ; if in different planes, it is termed a space frame. 
For the present we shall deal only with plane frames. 

A framed structure is designed so that, as far as possible, 
there are only pulling and thrusting forces, causing tension 
and compression stresses respectively, in its members, bending 
actions being obviated. In Continental and American practice 
it is common to make framed structures — or trusses as they 
are called — ^pin- jointed, but in British practice the joints are 
nearly always riveted. In either case the forces in the members, 
or stresses as they are usually rather erroneously called, are 
calculated as if the joints were pinned; these joints are often 
called nodes. 

Kinds of Framed Structures. A framed structure may 
be one of three kinds, viz. deficient or under-firm, perfect or 
firm, and redundant or over-firm. 

A deficient or under-firm frame is one which has not sufficient 
bars to keep it in equilibrium for all systems of loading. Such a 
frame is shown in Fig. 95 (1). For certain values of the forces 
acting on it, the frame would be in equilibrium, but it would 
collapse if the forces were changed. 

A perfect or firm frame is one which has a sufficient number 
of bars — and no more — to keep it in equilibrium for all systems 
of loading. Such a frame is shown at (2) in the figure. 

A redundant or over-firm frame is one which has more bars 
than are necessary to keep it in equilibrium for all systems of 
loading. Such a frame is shown at (3) in the figure. 



OH. XI] THE FORCES IN FRAMED STRUCTURES 175 

Objections to Deficient and Redundant Frames. If a 
deficient frame is actually pin- jointed, it is in unstable equilibrium ; 
if its joints are riveted, then its stability depends on the stiffness 
of the joints and its members are subjected to bending stresses 
which it is the object of the framework to avoid. 




Fig. 95. 

Redundant frames have the following disadvantages: 

(1) Any stress in one member caused by bad fitting or change 

of temperature causes stresses in all the other members. 

(2) The stresses in the members cannot be calculated by 

any simple mathematical or graphical process. 

Such frames are sometimes called " statically indeterminate." 

Semi-member or Counterbraced Frames. Some frames, which 
have the appearance of redundant frames, act as perfect frames 
and may be treated as such. Fig. 95 (4) shows such a frame. 
There are two diagonal bars BD and ACy but each can act in 
tension only, so that if the loading is such as would tend to put 
one of the diagonals, say .40, in compression, such diagonal 
would go out of action and the frame would act as if BD were 
the only diagonal. 

The diagonals AC and BD are called semi-members or counter - 
braces and are commonly used in practice, especially in the centre 



176 THE FORCES IN FRAMED STRUCTURES [ch. 

panels of railway bridge trusses in which the crossing of the 
load causes a reversal of the stress in the diagonals. 

Relation between Bars and Nodes in a Perfect or Firm 
Frame. Consider a firm frame such as is shown at (2) in Fig. 95. 

The first bar DC has 2 nodes. 

It requires two more bars AD and AG to produce the next 
node -4, and so on. 

Therefore, if there are n nodes, 2 of them go to the first bar 
and the remaining (n — 2) require 2 (n — 2) bars. 

.*. Total number of bars = 2 (ti — 2) x 1 = 2n — Z. There- 
fore, in a perfect or firm frame the number of bars is equal to 
twice the number of nodes minus 3. 

If the number of bars is more than this, the frame is redundant ; 
if less, the frame is deficient. 

The student should test this relation with the framed structures 
shown in the following figures. 




Fig. 96. 

The converse of the above statement does not hold. The 
number of bars might be = 2w — 3, and yet the frame might 
not be perfect. 

Fig. 96 gives an example of this. In this case the number 
of nodes is 12 and the number of bars 21, so that this fulfils 
the above condition, although it is not a perfect frame. 

Ties and Struts. If the force in a member of a structure 
is a pull, the member is called a tie ; if the force is a thrust, the 
member is called a strut. The force in a bar transmitted from 
a certain node wUl be a pull when the arrow-head points away 
from the node, and a thrust when it points towards it. 

It is desirable to distinguish between the ties and the struts 
in the drawing of a framed structure. This can be done in any 
of the following ways: 

(1) By drawing the struts in thicker lines than the ties. 

(2) By drawing a short single line across the ties and a 

double line across the struts, e.g., | and ||. 



XI] THE FORCES IN FRAMED STRUCTURES 177 

(3) By indicating the struts with a plus sign and the ties 
with a minus sign. 

Loading of Framed Structures. Framed structures must 
always be taken as loaded at the nodes only. If a given bar 
is loaded between the nodes, then it acts as a beam and distributes 
to the nodes at each end the reaction of the beam. 

Curved Members in Framed Structures. In some cases the 
members or bars of a framework are curved. For obtaining the 
forces in the bars (not really the stresses, although this term 
is most often used), we replace the curved bars by straight 
ones; but it must be carefully remembered that such bars 
wiU actually be subjected also to bending stresses which must 
be allowed for in design. 

Determination of force in a Framed Structure. Since a 
pin- joint is provided at each node, the forces in the bars meeting 
in such a node and the external forces must be in equilibrium. 

Before the forces in a frame can be determined, the whole 
of the external forces, including the reactions at the supports, 
must be known. 7 

Example of simple Roof-Truss. Take for example the simple 
roof -truss shown in Fig. 97. The external forces acting are 
the weight W = 2000 lbs. at C and the reactions R^, Rb at A 
and B. We assume that the truss simply rests on its supports 
so that these reactions are vertical. Since the frame looks the 
same whether viewed from the front or the back, or "from 
considerations of symmetry" as this is usually expressed, the 
reactions must be equal to each other, and for the equilibrium 
of the structure as a whole, their sum must be 2000 lbs. 

.-. R^ = Rb=^ 1000 lbs. 

Now consider the node C There are three forces acting 
there, viz., W vertically downwards and forces f^ct Jbc iii 
AC and GB ; and these forces must be represented by a triangle. 

.*. draw 1, 2 to represent W ~ 2000 lbs. to a convenient 
scale and draw 1, 3 parallel to CA and 2, 3 parallel to GB to 
intersect at 3. Then 2, 3 = Jbc ^iid 3, 1 = Saq- C)n scaling 
these off, we find /t^ =/^(7 = 22401bs. (Thrust, because the 
arrow-head points towards the node). 

A.M. 12 



178 THE FORCES IN FRAMED STRUCTURES [c5h. 

Next consider the node A. The forces acting are the reaction 
i?^ vertically upwards and fcAy /ba in GA and BA. 

:. draw 4, 1 to represent Rj^ = 1000 lbs. and draw 4, 3 
parallel to BA and 1, 3 parallel to CA to intersect in 3. Scaling 
these off we find /ba = 3, 4 = 1000 lbs. (Pull, because the arrow- 
head points away from the node), fcA = 2240 lbs. (Thrust, as 
before). 



W = 2000 lbs. 




Rg^lOOO Lbs. 



Kg. 97. Simple Roof-Truss. 

It will be noted that the force transmitted to any bar from 
the nodes at its two ends must be equal and opposite ; otherwise 
that bar would not be in equilibrium. 

The Reciprocal Figure Construction. The reciprocal figure 
construction is an extension, devised by Clerk Maxwell, of the 
method that we have just explained ; the various vector polygons 
for each node are combined in one diagram. We will first explain 
the construction with reference to the roof -truss shown in Fig. 98. 
We will take the vertical loads on the nodes as equal, the reactions 
then being equal and vertical. 

To commence the reciprocal figure set down lengths 1, 2; 
2, 3; etc., on a vertical line to represent the forces, to some 
convenient scale, the reaction 4, 5 being equal to half the total 
load, and giving the point 5 as shown. At the left-hand end 



XI] THE FORCES IN FRAMED STRUCTURES 179 

of the truss three lines meet, viz., 5, 1 ; 1, ^ ; A, 5. On the 
reciprocal figure we require a corresponding triangle, so draw 
1, a parallel to 1, ^, and 5, a parallel to 5, -4, their point 
of intersection determining the point a on the reciprocal figure. 
From a draw ab parallel to AB, and 26 parallel to 2B, thus 
obtaining the point b; then be parallel to BC, and 5c parallel 
to 5(7, thus obtaining the point c, and so on. 

Frame diagram 

IW 




Reciprocal figure 
Fig. 98. Roof-Trass. 

To serve as a check on the accuracy of the drawing, the line 
joining the last point e on the reciprocal figure to the point 5 
should be parallel to the bar E5 of the frame. 

Then the lengths of the lines of the reciprocal figure give — 
to the scale to which the loads were set down — the forces in 
the corresponding bars of the frame. 

To distinguish between Ties and Struts. To ascertain which 
members of a framework are ties and which are struts, the 
following method is adopted and can be applied for all systems 
of loading. 

Consider any one of the nodes of the truss at which the 
direction of one force is known, say the node X. Corresponding 
to this node we have the polygon 12bal on the reciprocal figure. 
The direction of the force 12 is known to be vertically downward, 

12—2 



180 THE FORCES IN FRAMED STRUCTURES [ch. 

so continue the arrow-heads in this direction round the polygon 
126ol. Now transfer the direction of these arrow-heads to the 
corresponding bars close to the given node. Then if the arrow- 
head on a given bar points towards the node, the bar is a strut ; 
and if it points away, the bar is a tie. In this way it is seen 
that the bars 1^, AB, and B2 are all struts. 

Now consider the node Y. Corresponding to this we have 
the polygon 5abc5. Since AB is a strut, the arrow-head at the 
node Y points towards the node, and so the arrow-heads go round 
the polygon in the direction a6, be, c5, 5a, as shown. Transferring 
these arrow-heads to the frame diagram, we see that the bars 
BC, 05, and 5 A are all ties. 

With practice one can tell by inspection in most cases whether 
a given bar is a strut or a tie by the following rule : If, on imagin- 
ing the given bar cut through, the forces would tend to increase 
its length, such bar is a tie; if the forces would tend to 
decrease its length, the bar is a strut. 

Example of Warren Girder with Uneven Loading. As a 
further example of reciprocal figures, take the example of the 
Warren girder loaded as shown in Fig. 99. 




Fig. 99. Warren Girder. 

Unless it is definitely stated to the contrary we can always 
take it that in framed girders the " panels" or " bays" as they axe 



XI] THE FORCES IN FRAMED STRUCTURES 181 

called are of equal length, i.e. in this case the span is divided into 
six equal parts. 

We must first find the reactions before we can proceed with 
the reciprocal figure. Taking moments about X we have 

i2y X 60 = 1 X 10 + 1 X 20 + 1 X 30 + 2 X 15 + 5 X 45 = 315 ; 

.*. Ry = %V- = ^*25 tons ; 

.-. i?j: = 10 - 5-25 = 4-75 tons. 

Choosing a suitable force scale, we set down the vertical forces 
in order, i.e. first set down 1, 2 and 2, 3 to represent 2 and 
5 tons respectively ; next set up 3, 4 to represent the reaction 
of 5 "25 tons ; and then set down 4, 5 ; 5, 6 ; and 6, 7 to represent 
1 ton each, the length 7, 1 checking back to give the reaction 
4*75 tons. We now proceed as before, drawing la parallel 
to 1^, and 7a parallel to lA ; then ab and 16 parallel respectively 
to AB and 15, and so on, the reciprocal figure coming as shown, 
and 13 coming parallel to L3, and thus serving as a check on the 
drawing. 

In cases of complicated frames where some difficulty is 
experienced of getting the last line to check, it is well to start 
the reciprocal figure from each end of the frame, the errors 
being in this way minimised. 

Figs. 100, 101 show the reciprocal figures for two other common 
forms; the student should work these as an exercise; with a 
Little practice it wiU be found that these diagrams can be drawn 
without difficulty and they have the great advantage that the 
closing line forms a check on the work. 

There need be no hard and fast rule as to the end from which 
we commence the figure. In Fig. 101, for instance, we have 
commenced from the right-hand side whereas in the other cases 
we have commenced from the left-hand side. When two points 
on a reciprocal diagram coincide, and the line joining them 
corresponds to a bar in the frame, the force in the corresponding 
bar is zero. Thus in Fig. 101 the stress in FF' is zero and so 
we have drawn the bar in dotted lines; the other bar shown 
dotted being a semi-member or counterbrace (see p. 175). The 
bar FF' is not really redundant because if the loading were 
altered, there would be a stress in it. 

Stresses in Framed Structures by Moments. The stresses 
in framed structures can also be found by the following method, 



182 THE FORCES IN FRAMED STRUCTURES [ch. 





Fig. 100. Crescent Roof-Truss. 




Fig. 101. 



XI] 



THE FORCES IN FRAMED STRUCTURES 183 



which is called the method of Moments or Sections, or sometimes 
Ritter's method. The method consists in imagining one bar 
to be cut through and in finding the point about which the 
structure tends to collapse. Consider for example the simple 
roof-truss shown in Fig. 102 and suppose that the bar AB is 




R^'lOOO Rb»iooo 

Fig. 102. The method of Moments. 

cut through ; the ends A and B will then move outwards some- 
what as indicated in dotted lines and the bars AC, BG will 
pivot about the point G. Therefore the force /^^ in AB must 
be the force which prevents this pivoting movement and its 
moment about the point G must be equal to the moment of 
either reaction about G, 



I.e. 



. 1000 X 8 „^^^ „ 
Sab= -. = 2000 lbs. 



In using this method it is best to regard one side of the 
structure as remaining fixed and the other side as moving under 
the action of the forces upon it ; in the present example therefore 
we regard BG as fixed and -4C as moving upward, pivoting 
about G. 

Now let us find the force f^c\xs.AG. 11 AG were cut through 
the weight W would fall down and the bar GB would pivot 
about the point J5. The only force tending to cause this collapse 
is W whose moment about B = 2000 x 8, 

.*. fAcxy = 2000 X 8, 

- _ 2000 X 8 _ 2000 X 8 
•'^^" y ~ 716 
= 2240 lbs. nearly. 



/ 



184 THE FORCES IN FRAMED STRUCTURES [ch. 

Experiment upon Model Roof-Truss. A simple experimental form 
of roof-truss similar to that which we have already considered is shown in 
Fig. 103. The struts AC^ CB (called rafters) are each formed of two parts 
sliding one within the other and connected by compression spring balances 
which will measure the forces in these bars. The tie-bar AB consists of two 
pieces of wire or string connected by an ordinary spring balance which will 
measure the pull in it. 




777^777777777777777777777777777777777777' 
Fig. 103. 

A more accurate form of experiment consists in making a model frame 
in iron and measuring the small changes in length of the various members 
by means of a very sensitive instrument called the extensometer (see p. 147), 
from the readings of which the forces acting in the various bars can be readily 
calculated. 

Forces in Tripods and Shear Legs. In these cases we proceed 
as follows : 

Draw the structure in plan and elevation, and let W be the 
load at A, Fig. 104, AB being the back leg and AD, AE the fore 




Fig. 104. 



XI] THE FORCES IN FRAMED STRUCTURES 185 

legs. Resolve W down AB and down the 'plane of the other two 
legs, i.e., set out ah equal to W and draw he parallel to AB^ and 
a^ parallel to AC^ then he represents the force in AB. Now 
swing the shear legs down horizontally in order to get A.flE, the 
true shape of the triangle ADE, then setting out ac horizontally 
and drawing ad and cd parallel to EA^, DA^ respectively, we get 
the forces in the fore legs. 



SUIVOIARY OF CHAPTER XI. 

Frames may be deficient, firm or redundant, but only in the case 
of firm frames can the forces in the members be found by simple 
methods. 

Before the forces in the various members of a frame can be 
calculated, all the forces acting upon it, including the reactions, 
must be known. 

At each node, the forces in the members meeting there and the 
external forces acting there must be in equilibrium and so must 
form a closed polygon. 

Members subjected to pulling forces are called ties and those 
subjected to thrusting forces are called struts. The reciprocal 
fig\u*e construction forms an automatic check upon its accuracy 
because the closing liue must be parallel to the corresponding bar 
in the frame. 

The method of moments enables the force in any particular 
bar to be calculated and by applying the method to a convenient 
bar, preferably near the middle of the structure, we get a useful 
check upon the accuracy of the grapliical construction. 



186 THE FORCES IN FRAMED STRUCTURES [ch. 



EXERCISES. XI. 

1. Find the forces in the members of the derrick crane shown 
in Fig. XI a. 



10 Tons 




1 

1( 


1 




^ 




>^ 




1 




V 


^""^^ 


B 


. — a/ 




c \ 






' 1 


■ 



Fig. XI a. 



ITon 
Fig. XI c. 



ITon 



2. Find the forces in the members of the framed structure 
shown in Fig. XI h and check your result for BB' by moments. 




Fig. XI 6. 



3. A Warren girder of length 100 feet is divided into 5 bays 
on the lower flange, the length of the inclined braces being 20 feet. 
If loads of 30 tons are carried at the lower nodes 20 and 40 feet 
from one end find the forces in the members. 

4. Find the forces in the members of the cantilever truss shown 
in Fig. XI c, which projects from a wall AB. [Note BC = CD 
and AE = EF ^ FD.] 



XI] 



THE FORCES IN FRAMED STRUCTURES 187 



5. Draw the reciprocal figure for the roof -truss shown in Fig. XI d 
given that AB = BC ^CB' = B'A' ; AD = A'D' = CD = CD' and 
scale off the force in DD\ 



2Tona 



3 Tons 




Fig. Xld, 

6. A load of 7 tons is siispended from a tripod the legs of which 
are of equal length and incUned at 60° to the horizontal. Find the 
thrust on each leg. 

7. A pair of shear legs (Fig. XI e) make an angle of 20° with 
each other and their plane makes an angle of 60° with the horizontal. 
The back stay is at an angle of 30° to the horizontal. Find the 
force in each leg and in the stay when supporting a load of 10 tons. 




Fig. XI e. 



CHAPTER XII 



BEAMS AND GIRDERS 



We shall get a good preliminary idea of the stresses occurring 
in beams by considering a model devised by Prof. Perry. Suppose 
that a beam fixed at one end carries a weight W (Fig. 105) at the 





Te n a J o n 



Fig. 105. Stresses in Beams. 

other end, and that it is cut through at a certain section. Then 
the right-hand portion can be kept in equilibrium by attaching 
a rope to the top and passing over a pulley, a weight W being 
attached to the other end of the rope, and by placing a block 
B at the lower portion of the section and a chain A at the upper 
portion. Then the pull in the rope overcomes what is called the 
shearing force ; and the block B carries a compressive force 0, 
and the chain A carries a tensile force T. Since these are the 
only horizontal forces, they must be equal and opposite, and thus 
form a couple. Then the moment of this couple must be equal 



CH.xn] BEAMS AND GIRDERS 189 

and opposite to the couple due to the loading, which is called 
the bending moment. 

In the actual beam, owing to the deflection which takes place, 
the material on one side of the beam will be stretched, and the 
material on the other side will be compressed, so that at some 
point between the two sides the material will not be strained at 
all, and the axis in the section of the beam at which there is no 
strain is called the neutral axis. To tell whether the top or the 
bottom is in tension we consider the deflected form which the 
beam will take up and note that the tension edge is always on the 
outside of the bend whereas the compression edge is on the inside. 

Shearing Force and Bending Moment. The actual calcula- 
tion of the stresses in a beam is beyond our present scope but 
such calculations depend upon the quantities called Shearing Force 
and Bending Moment which are quite simple to understand and 
with which we will now deal at some length. 

Definitions. The shearing force at any point along the span 
of a beam is the algebraic sum of all the perpendicular forces 
acting on the portion of the beam to the right or to the left of 
that point. 

The bending moment at any point along the span of a beam is 
the algebraic sum of the moments about that point of all the 
forces acting on the portion of the beam to the right or to the 
left of that point. 

As the beam is in equilibrium under the forces acting on it, at 
any point the algebraic sum of the forces, and of the moments of 
the forces about the point, acting on both sides must be nothing ; 
so that we shall get the same numerical values for the shearing 
force and bending moment from whichever side we consider them, 
but they will be opposite in sign. We will, wherever possible, 
always consider the shearing force and bending moment of the 
forces to the right of the section, and we will take an upward 
shearing force and a clockwise bending moment as positive, the 
downward and anti-clockwise being taken as negative. 

Bending Moment and Shearing Force Diagrams. If the 
bending moment and shearing force at every point of the span 
be plotted against the span and the points thus obtained be 
joined up, we shall get two diagrams called the Bending Moment 
(b.m.) and Shear diagrams, and from these diagrams the values 



190 



BEAMS AND GIRDERS 



[CH. 



of these quantities can be read off at any point of the span. We 
will examine the forms of these diagrams for various kinds of load- 
ing and for various ways of supporting the beam, and will first con- 
sider beams with fixed loads. We will use Mp and Sp to represent 
respectively the bending moment and shearing force at a point P. 
We will restrict our consideration to loads which are fixed in 
position as opposed to those which may roll from one position to 
another. 



A 
I 



<;"«?■ 



t 

vvi 



i 



-^L 



>.A 



•X' 




Isolated Load 
Fig. 106. Cantilevers. 



A. Cantilevers, i.e. beams fixed at one end and free at the 
other, the loads being all at right 
angles to the length of the beam. 

Case 1. Cantileveb with 
One Isolated Load. Let a canti- 
lever, fixed at the end 5, Fig. 106, 
carry an isolated load W at the 
point u4, at distance I from B. 
Consider any point P at distance x 
from A. 

Then we have 

Bp=W, 

This is constant throughout the 
span. 

.*. Shear diagram is a rectangle of height TF. 

Again Mp = TF x a;. 

This is proportional to x. 

:. B.M. diagram is a triangle 
whose maximum ordinate is Tf Z, 
this being the bending moment at 
the point B. 

Case 2. Cantilever with 
Uniform Load. Let a uniformly 
distributed load of 'p tons per foot 
run be carried by a cantilever AB 
of span Z, Fig. 107. Consider a 
point P at distance x from the 
free end A. Then 

Bp = load on AP 
-= 'px. 



^XDCQQQQpD 




Un I jo rm L oaci. 
Fig. 107. Cantilevers. 



Xll] 



BEAMS AND GIRDERS 



191 



This is proportional to ic, and therefore the shear diagram is a 
triangle, the maximum shear occurring at the end B, and being 
equal to pi ov W^iiW i& the total load on the cantilever. 

Mp — Moment of load 'px about P 



X 

= pxx ^ 

" ~2~' 

This is proportional to x^, and 
therefore the b.m. diagram will 
be a parabola with vertex at A. 
The maximum b.m. will be equal 

Wl 

2 



trcTTY^C)rrD_ 



. pi^ 

to ^ or 



and occurs at B. 




Afim 



Unjorm and isolated! Loocfs 

Fig. 108. Combined loading on 
Cantilevers. 



Case 3. Cantilever with 
Isolated Load and Uniform 
Load. Since the b.m. and shear 
at any point are defined as the 
sum of the moments and the 
forces to the right of that point, 
it follows that the b.m. and shear 
diagrams for a number of loads 
can be obtained by adding to- 
gether the diagrams for the 
separate loads. 

In adding together two diagrams we first draw the separate 
diagrams and then make diagrams whose ordinates at each point 
are the sums of the ordinates of the separate diagrams at the 
same point. 

Case 4. Cantilever with Irregular Load System. — 
Graphical Method. Suppose a number of loads 0, 1, 1,2, and 
so on, Fig. 109, act on a cantilever. To obtain the shear and 
B.M. diagrams set down 0, 1, 1,2, 2,3, etc., down a vector line 
0, 5 to represent the forces to some convenient scale, and take 
a pole P at some convenient distance p from the vector 
line 0,5 and join P to each of the points to 5 on the 
vector line. 

Now across the lines of the forces draw ag parallel to PO; 



192 



BEAMS AND GIRDERS 



[CH. 



across space 1 draw ab parallel to PI ; across space 2 draw be 
parallel to P2, and so on until the point / is reached. 
Then abcdefg is the b.m. diagram. 



s oO 3 020 1 Q ) 




Fig. 109. Graphical method for Cantilevers. 



To obtain the shear diagram, project the points 0-5 on the 
vector line across their corresponding spaces, the line through the 
point being drawn right across the span, the stepped figure thus 
obtained being the shear diagram. 

Proof. Consider any point P along the span, and produce 
ab and be to cut the corresponding ordinate Pj P2 ^f the link 
polygon at 6' and c' respectively. 

Now consider the triangles aPjp' and POl. 

They are similar, and as the bases of similar triangles are 
proportional to their heights, we have 

P^b^ ^ aPi 
0, l~ p ' 

:.p X P^b' = 0, 1 X aP^. 

But 0, 1 X aP^ = Moment of force 0, 1 about P. 
/. p X PJ)' = Moment of force 0, 1 about P. 



xn] BEAMS AND GIRDERS 193 

Similarly it follows that 

p X h'd = Moment of force 1, 2 about P, 
and 'p X c'F^ = Moment of force 2, 3 about P. 

.-. we see that 'p x PiP2= p (Pi^' + 6'c' + cTg) 

= Moment of all forces to left of P about P 

:. since p is a constant quantity, it follows that the ordinates 
of the link polygon represent the bending moments at the corre- 
sponding points of the beam. 

Now consider the shear S at P. The total force to the left of 
P is 0, 1 4- 1, 2 + 2, 3 = 0, 3, and this is obviously the value given 
on the shear diagram. 

Scales. In all graphical constructions it is extremely 
important to state clearly the scales to which the various 
quantities are plotted, and to see that such scales are con- 
venient for reading off. 

Let the space scale be 

1 in. = a; feet 

and the load scale on the vector line 

1 in. = 2/ tons, 

and let the polar distance be p actual inches. 

Then the scale to which the bending moments can be read 
off is 1 in. = ^ X a; X 2/ tons-ft. 

p should thus be chosen so as to make this a convenient 
round number. 

To take a numerical example, suppose the space scale is 
1 in. = 4 ft. and the load scale is 1 in. = 2 tons, then if p is taken 
as 2i ins. the b.m. scale will be 1 in. = 4 x 2 x 2J = 20 tons-ft. 

If p had been taken 2 ins. the b.m. scale would be 
1 in. = 16 tons-ft. which would not be nearly such a convenient 
scale. 

B. Simply Supported Beams, i.e. beams simply resting on 
two supports, the loading all being at right angles to the length 
of the beam. Unless it is definitely stated to the contrary, we 
will always take it that the supports are at the ends of the beam. 

In simply supported beams the forces acting are the loads 
and the reactions at the supports, the sum of the reactions being 

A. u. 13 



194 



BEAMS AND GIRDERS 



[CH. 



equal to the total load, and their values being obtained by means 
of moments as explained in Chapter II. As the ends are freely 
supported, there can be no bending moment at either end. 
We will now consider the following standard cases : 

Case 1. Isolated Load in any Position. Let a load W 
be supported at a point C, Fig. 110, on a beam AB of span Z, the 
distances of the point C from B 
and A being b and a respectively. 

To obtain the reaction Bb at 
B take moments round A, 



Then 

Rb xl 

Bb 

Similarly 



A^f 



Wxa, 

Wxa 

I 

Wxb 
I 






Now consider a point P be- 
tween B and C. 



a 



Kh 



^ 



I 



K-a; 



.R. 



V///A/y//A 



Shear Diagram 




B 






8 -B +^°- 

Op •— Hb J , 



/so/aTed Load 
Fig. 110. 



.*. between B and C the shear diagram is represented by a 
rectangle of height 

Wa 



Now take a point between C and A. 
Shear = Bb-W 

I 



f-w^w{^)^ 



= -R 



A 9 



height 



Shear between C and A is represented by a rectangle of 



In the case of the cantilever there was no need to distinguish 
between positive and negative shear because there was no change 
in direction of the shear ; but in the present case there is a change 
in direction, and so we will use the rule given on p. 189. 



XII] BEAMS AND GIRDERS 196 

Now considering the bending moment, 

, , „ W .a.x 
Mp = B^x X = J — , 

This is proportional to x, and therefore the b.m. diagram 
between B and C will be a triangle, the b.m. at (7 being equal to 

Wab _ Wa jl - a) 
I ~ I ' 

If P were between C and A and at distance x' from A we should 
have 

Mp ^Bsil- x') -W {l-x' ^b) 

= Bs.l'-Bs.x'-Wl+ Wx' + Wh 
-= a;' (IT - Bs) + TF6 - 2 (PF - B^) 
= B^.x' + Wb- lBj_ 

Whr' 

= iip j^wb-Wb 

_ Wbx' 
~ I ' 

This is proportional to x\ and therefore the b.m. diagram 
between A and C is also a triangle, the whole diagram then 
being as shown in the figure. 

Case 2. Isolated Load at Centre. This is a special case 

of the preceding one, in which a = 6 = ^. 

W 
Each reaction is now equal to -^ and the maximum b.m. 

TTx -X - 
_ 2 2 Wl 

I - 4 • 

Case 3. Uniform Load over Whole Span. Let a uniform 
load of p tons per ft. run cover the whole span ABy Fig. Ill, and 
consider a point G at distance x from B. 

In this case the two reactions will, from symmetry, be equal, 

and each have the value ^ or -^ . 



Theti So^Rb-P^^pU-^* 



13—2 



196 



BEAMS AND GIRDERS 



[oh. 



This is a linear relation, therefore the shear diagram will be 

pi 
a triangle as shown, having values ± ^ at the ends and changing 

sign at the centre. 

Now consider the bending mo 
ment. 



a^- 



nnnncYTYYYYTY- 



z 



Mq = RsX X — pxx ^ 






plx 
"2" 



"2" ~ 2 ^ 



x^). 



This depends on x\ and there- 
fore the B.M. diagram will be a 
parabola. 

The maximum b.m. will occur 

at the centre, i.e. when x = ■^, 



p Torn f>er Ft 



I 



c*=FP^% 






t)oroho!m 




Then maximum b.m. 

=![(¥)-©■] 



Uniform load over mi hole, s^xin 
Fig. 111. 



~2V2 4:) 



p P pl^ Wl 
— ^ — = =- — or — 
2 4 8 8 



Case 4. Ireegulak Load. — Graphical Construction. 
Let a number of loads PFj, TFg, W^, and W^ be placed anjrwhere 
along a span AB, Fig. 112. Number the spaces between the loads 
and set down 0, 1 ; 1, 2; 2, 3; 3, 4 as a vertical vector line to 
represent the loads to some convenient scale, and in any position 
take a point P at a suitable polar distance p from the vector 
line, and join PO, PI, P2, etc. 

Across space then draw ab parallel to PO; across space 1 
draw be parallel to PI and so on until ef is reached, this being 
parallel to P4. 

Join a/, then the figure a, b, c, d, e, /, a will give the b.m. 
diagram for the given load system. 

Now draw Px parallel to a/, the closing link of the link 
polygon ; then on the vector line, 4:X = R^ and xO = i?^ . 

To draw the shear diagram, draw a horizontal line through 
X right across the span : this gives the base line for shear. Now 
project the point horizontally across space 0; project point 



xn] 



BEAMS AND GIRDERS 



197 



1 across space 1 and so on, the stepped diagram thus obtained 
being the shear diagram. 

If the first and last links are produced to meet at F, then as 
we proved on p. 28, the resultant load acts through Y. 

Peoop. By reasoning by similar triangles as for the cantilever 
we can prove that 4a; = B^ and icO = -B^ . 



Hi 



O g n^3 O 4 




Fig. 112. Graphical construction for simply supported beam. 



Now consider any point R along the span. 

= 4a; - 3, 4 = Sx, 

but the ordinate S of the shear diagram is equal to Sx, and 
therefore the stepped figure gives the correct shearing force at 
any point. 



198 BEAMS AND GIRDERS [ch. 

Let the vertical through R cut the b.m. diagram in -Rii?2 and 
fe produced in €2. 

Then by exactly similar reasoning as before 



!Rje2 = 



JXnSn 



Moment of Rb about R 

P 

Moment of W^ about R 



^2-2 p 



Moment of Rn — Moment of Wa about R 



V 



Mn^ 



V 
:. Mr= px R^Rz* 

:. the ordinate of the b.m. diagram represents the b.m. at 
any point. 

Scales. As in the case of the cantilever (page 193), if 
I*' = X feet is the space scale and I'' = y tons is the force scale, 
and if the polar distance is p actual inches, then the vertical 
ordinates of the b.m. diagram represent the bending moment to 
a scale V = p x x x y tons ft. 

Note. — In this construction the bending moment R1R2 is 
measured vertically and not at right angles to the closing line af. 

The above construction is a special case of the link and vector 
polygon construction described on p. 28. 

Numerical Examples. (1) A freely supported beam of 20 ft. 
span carries a uniformly distributed load of 5 tons, and isolated 
loads of 3 and 2 tons, at distances respectively of 4 and 5 ft. from 
the ends {see Fig. 113). 

We have first to get the reactions i2^ and R^, 

Take moments round B. 

72^x20 = 5x10 + 3x16 + 2x6 
= 60+48+ 10= 108, 

D 108 ^ . ^ 

.-. i2^ = 10 - 6-4 = 4-6 tons. 
The shear diagram is then as shown in the figure, the amounts 



XIl] 



BEAMS AND GIRDERS 



199 



of the steps being equal to the isolated loads. The point at 
which the shear is nothing is found as follows: 
Let it be at distance x from B. Then 

^^ = = 7^5 - 2 - i) . a: 
5a; 



= 4-6 - 2 - 



20 



2-6- 



X 

4' 



X 



:. I = 26, 



X = 10-4 feet. 



ZTons 



2 Tons 



4*6 T 




5«4T 



Fig. 113. 

The B.M. at this point will be a maximum, and will be equal to 

1 10-42 
M^ = RsX 10-4 - 2 (10-4 - 5) - ^ . ^ 

= 47-84 - 10-8 - 13-52 
= 23-52 tons-ft. 



200 BEAMS AND GIRDERS [ch. 

The B.M. diagram will consist of a parabola for the uniformly 
distributed load, the maximum ordinate of which is equal to 

6 X 20 , o e i. ti. 

— ^ — =■ 12-6 tons-ft 

o 

The B.M. diagram for each of the isolated loads will be a triangle, 
the respective heights being 

3x4x16 -_. ,. ,2x5x15 __. ., 
^ = 96 tons-ft. and ^a = ^'^ tons-ft. 

Combining these three figures we get the b.m. diagram shown 
on the figure, and on scaling off the maximum ordinate it will be 
foimd to be 23*5 tons-ft. 

Note. — In all constructions where diagrams are going to be 
added together, such diagrams must of course be drawn to the 
same scale. 

(2) A certain joist used as a cantilever weighs 18 lbs. per foot, 
and the maximum B.M. which it can safely carry is 63-56 tons-ins. 
Find how long the span may he for the cantilever to be able to sustain 
safely its own weight. 

We have for a cantilever 

Max. B.M. = ^ ; 

Now p = 18 lbs. per ft.; .*. max. b.m. = 63-56 tons-ins. 

63-56 X 2240 » *. . •* , • • r . 

= Yn Ibs.-ft. ; :, 111 is span m feet we have 

2 X 63-56 X 2240 

* - i2^n:8 - *^^^' 

I = n/13I8 = 36-3 feet. 

(3) A beam 20 ft. span carries loads of I, J, 1 and 2 tons, as 
shoum on Fig. 15. Determine graphically the maximum B.M. 

Draw the b.m. curve by the link and vector polygon construc- 
tion as shown in Fig. 22. Take the space scale I'' = 4 ft. ; the 
load scale 1" = 2 tons ; and the polar distance 1 J inches. The 
maximum ordinate of the b.m. curve will then be found to be 
1-09 inches. The scale of this will be 1" = 1 J x 4 x 2 = 10 tons-ft. 
.*. Maximum b.m. = 10-9 tons-ft. 



XII] BEAMS AND GIRDERS 201 



SUMMARY OF CHAPTER XII. 

Neutral axis of a beam is the line in the cross section which 
receives no stress or strain. 

The stresses in a beam are tensile on one side of the neutral 
axis and compressive on the other, the resultant forming a couple 
whose moment must be equal to the bending moment. 

The shearing force at any point is the algebraic sum of all the 
perpendicular forces acting on the portion of the beam to the right 
or left of that point. 

The bending moment at any point is the algebraic sxan. of the 
moments about that point of all the forces acting to the right or 
left of that point. 

Bending moment and shear diagrams can be drawn for standard 
methods of loading and fixing the ends of the beam; for two or 
more loadings occurring together, the separate diagrams are added 
together. 

For a simply supported beam of span I with an isolated load W 
at distance a from one end 

Max DM- ^"^^"^^ 

Wl 

For uniform load Max. b.m. =-^ . 

o 

Shear and bending moment diagrams for any loading may be 
drawn by means of the link and vector polygon construction. 



EXERCISES. XII. 

1. A beam AB 15 ft. long is fixed at A and free at the other 
end. A weight of 80 lbs. is placed at the end B, and a weight of 
100 lbs. in the middle of the beam. Find the b.m. and s.f. at dis- 
tances of 3, 6 and 9 ft. from the fixed end. 

2. A beam 150 ft. span is uniformly loaded with 2 tons per 
ft. run. Calculate the b.m. and s.r. at every 10 ft. of its span and 
draw the curves of b.m. and s.f. 

3. A girder supported at both ends is 60 ft. span and carries 
a uniformly distributed load of 200 lbs. per ft. rim. Find the b.m. 
and S.F. at the centre and at points 15 ft. from the ends. - 



202 BEAMS AND GIRDERS [ch. xn 

4. A cantilever is 36 ft. long and is uniformly loaded with 
150 lbs. per ft. run. Find the b.m. and s.f. at the fixed end and 
at points 25 ft. and 15 ft. from the fixed end. 

5. If in the last question an additional load of 750 lbs. is placed 
in the centre of the beam find the magnitude of the b.m. and s.f. 
at the points mentioned. 

6. A beam 45 ft. span supported at the ends carries a weight 
of 6 tons 15 ft. from one end. Find the b.m. and s.f. at the centre 
and also at 5 ft. from each end. Draw a diagram of b.m. and s.f. 
to scale. 

7. A beam is loaded as shown (see Fig. XII a). Find the reactions 
at the supports, also the b.m. and s.f. at each quarter span, C, D 
and E being the points at quarter distance each along the beam. 



2 3 4 5Tons 



'm Tc Di Te — : — W 



B 



81 • 15' ' 7' • r '5^ 

Fig. XII o. 

8. A truck weighing 10 tons is carried on four wheels. The 
distance apart of the centres of rails is 5 ft. and of the axle boxes 
is 6 ft. 4 ins. Find the b.m. and s.f. at points on the axle 4 ins. 
apart. Draw diagrams. 

9. A imiform beam, weighing 180 lbs. per ft. run, 20 ft. long 
is supported at the ends. It carries also a load of 60 lbs. in the 
centre. Find the b.m. and s.f. at points 5 ft. and 8 ft. from one end. 

10. Find the b.m. and s.f. at the centre of a beam 50 ft. long 
weighing 500 lbs. per ft. run and carrying a load of 10,000 lbs. 
20 ft. from one end. Draw the b.m. and s.f. diagrams. 

11. A plank placed across an opening 12 ft. wide is broken 
by the bending eSect of 3 cwt. placed 3 ft. from one end. What 
is the greatest load which a man weighing 156 lbs. could safely 
carry across? 

12. Find the b.m. at the centre of a girder 40 ft. long supported 
at the ends and uniformly loaded with ^ ton per ft. run ; there being 
an additional load of 5 tons 8 ft. from one end. 

Find also the s.f. at the centre. 

13. A timber beam is 18 ft. between supports and is 12 ins. 
deep by 4 ins. broad. Draw curves of b.m. and s.f. produced by 
its own weight; giving numerical values at each quarter span. 
Weight of timber 48 lbs. per cubic ft. 



CHAPTER XIII 

CENTRE OF GRAVITY AND CENTROID 

The centre of gravity of a solid body or of a number of bodies 
is a point through which the resultant weight of the whole body 
or bodies may be considered to act. Every portion of a body is 
attracted towards the centre of the earth by a force called the 
weight of that portion, and in a body of reasonable size the 




Fig. 114. Centre of gravity. 

weights of all the parts will be parallel, so that the problem of 
finding the centre of gravity of a body resolves itself into that 
of finding the position of the resultant of a number of parallel 
forces, and this is solved, as we have already seen on p. 19, by 
the principle of moments. 

Let 1, 2, 3, 4, etc.. Fig. 114, represent a number of very small 
bodies in the same plane whose weights w;i, t^2» ^3» ^4» ^^' ^^^ ^* 
right angles to the plane of the paper, the bodies being so small 



204 CENTRE OF GRAVITY AND CENTROID [CH. 

that the weight of each may be considered as acting through their 
centre. Suppose that is their centre of gravity ; then is the 
point through which acts the resultant W of the parallel forces 
equal to the various weights. The magnitude W of this resultant 
will be the sum of the separate weights, i.e. 

W = Wi-\- W2-\- w^-\- w^-\- .... 

For convenience this is written (as on p. 20) 

W=^Sw^ (1). 

Now suppose that OX and OY are any two convenient lines 
which are not parallel ; it is usually most convenient to take them 
at right angles to each other. 

Then, remembering that the forces are acting at right angles 
to the plane of the paper we have, by taking moments about OY, 

Total moment about Y 

= My = w^x^ ■\- W2X2-\- w^x^ + w^x^ + . . . , 

i.e. My = Sw^x^ (2). 

But if the resultant weight acts through G we shall have by 
the principle of moments 

PF . Z = Sum of moments of the separate forces about OY 
= Sw^x^ ; 



(3) 



_ Uw^Xi 

"'Ew^ 

__ Sum of moments of separate weights about Y 
Sum of separate weights 

Similarly by taking moments about OX we shall have 

^=^ ^*) 

_ Sum of moments of separate weights about OX 
Sum of separate weights 

In this way we have fixed the distance of the centre of gravity 
from two given lines and so have found its exact position. 

Application to continuous bodies. We can apply these 
principles to the determination of the centre of gravity of con- 
tinuous bodies by imagining such bodies to be divided up into 




xni] CENTRE OF GRAVITY AND CENTROID 205 

a very large number of very small parts, as indicated in Fig. 115, 
and regarding the weight of each separate part as acting through 
its centre. The greater the number of parts, 
the more accurate will our calculation be ; as 
the number of parts becomes very great, how- 
ever, the calculation becomes very laborious 
and it is seldom made in this way but by means 
of a branch of mathematics called the calcultcs, 
which every engineering student should study 
if he wishes to understand easily the more ^^' ^^^* 

advanced portions of mechanics. 

The centre of gravity as the balance point. If a body 
balances about a point or a line then that point must be on the 
vertical line through the centre of gravity or the line must inter- 
sect that vertical line ; moreover, if a body be freely suspended 
by a string or wire the wire or string must pass through the 
centre of gravity. 

Take first the case of a body balanced about a point A, 
Fig. 116; there are only two forces acting upon the body, viz. 
the resultant weight W of the body and the upward pressure 
R at the support. Since the body is balanced, it must be in 
equilibrium under the action of these two forces and if two forces 
act upon a body and keep it in equiUbrium they must be equal and 
opposite. If they were not, there would be a resultant moment 
about some points and the body would start turning. 

The weight W acts vertically downwards, therefore the 
upward pressure R acts vertically upwards, so that the centre 
of gravity G must lie upon the vertical through the point of 
support A. 

By exactly similar reasoning in the case of a body suspended 
by a string attached at a point B, the only forces acting are the 
weight W of the body and the tension T in the string ; therefore 
the direction of the string must pass through the point G. 

Now we have seen that the sum of the moments of a number 
of forces about any point is equal to the moment of the resultant 
about the same point, the resultant weight of a body passes 
through the centre of gravity and a force has zero moment about 
a point in its line of action. We therefore deduce the very 
important rule that the sum of the moments about the centre of 



206 CENTRE OF GRAVITY AND CENTROID [ch. 

gravity of the weights of the separate portions of a body making up 
the whole body must be zero. It is clear for instance from Fig. 116 




Fig. 116. Centre of gravity as the balance point. 



that the moments of the portions to the left of G are anti-clock- 
wise, while those to the right are clockwise and therefore of 
opposite sign. 

Centre of gravity by inspection. The centre of gravity of 
a body which possesses a section of symmetry will always be in 
that section. By a section of symmetry is meant a section 
which will divide it into two exactly similar parts, which are 
"looking-glass pictures of each other.'* 

If a body has two different sections of symmetry, the centre 
of gravity will always be on the intersection of the two sections. 

Take for instance the cylinder represented in Fig. 117 which 
is assumed to be of the same material throughout. Three 
sections of symmetry are shown ; one vertical, cutting the cylinder 
along the centre of its length : one at right angles to this, also 
vertical and cutting through the centre of the two ends : and the 



xm] CENTRE OF GRAVITY AND CENTROID 207 

third horizontal, also cutting through the centres of the two ends. 
The centre of gravity G is at the intersection of the three sections. 




Fig. 117. 



As a proof of the statement that the centre of gravity must 
lie upon a section of symmetry, consider the body shown in 
Fig. 117 a of which XX is a section of symmetry. For the purpose 
of the argument we will suppose that the body is what is called 
a solid of revolution y i.e. it is a body such that all transverse 
sections such as FF are circles, in other words it is the kind of 
body that we could turn in a lathe, the axis of rotation being in 
the section XX. 

Now consider two equal portions, 
each of weight Wy opposite each other 
and at the same distance from XX ; 
the moment of one about XX is wx 
and of the other — wx so that the 
sum of the two moments about XX 
is zero. As the whole body might 
be divided up into similar neutraUsing 
portions, the total moment of the 
whole of the separate weights about 
XX must be zero; in other words 
XX must pass through the centre of 
gravity. It will be noted that in the 
case of the cylinder, G is what in 
ordinary language we should call the 
geometrical centre of the cylinder 

and in all similarly regular bodies the geometrical centre of 
the body is the same as the centre of gravity. But if the 




208 CENTRE OF GRAVITY AND CENTROID [ch. 

body be made of material of varying density or if there are 
blow -holes in it this will not be the case. 

Numerical Example. A uniform rod 24 inches long weighs 
10 lbs. and carries at its ends balls of 4 and 6 inches diameter 
weighing respectively 5 and 8 lbs. Where is the centre of 
gravity ? 



&. 



24" 



51b. 



12" 



X 



1 

10 lb. 

I 

Y23lb. 




Fig. 118. 



Referring to Fig. 118, the combined weight of 23 lbs. acts 
at the centre of gravity G, and since the separate bodies are 
symmetrical their weights act at their geometrical centres as 
shown. 

We may take moments about any convenient point; take 
for example the centre A of the left-hand ball. Then we 
have 

Moment of left-hand ball about .4 = 5x0 «= 

Moment of rod about -4 = 10 x 14 (clockwise) = + 140 

Moment of right-hand ball about 4 = 8 x 29 (clockwise) = + 232 

Total (in inch-lbs.) = -f 372 

This must be equal to the moment of the total weight of 23 lbs. 
about A, 



I.e. 



23a; = 372, 

X = -^ = 16-2 inches. 



As a check the student should solve the problem by taking 
moments about the centre B of the other ball. It should be noted 
that there is no need to choose the centre of one of the separate 
bodies for taking moments, although that usually makes the 



xiii] CENTRE OF GRAVITY AND CENTROID 



209 



calculation easier. For instance we might have chosen the point 
G of junction of the rod with the left-hand ball Then we have 

Moment about C of left-hand ball =6x2 (anti- 
clockwise) = — 10 

Moment about C of rod = 10 x 12 (clockwise) = + 120 

Moment about C of right-hand ball = 8 x 27 (clockwise) = + 216 

Total (in inch-lbs.) = -f- 326 

/. Dist. GC^^^ 14-2 inches. 
This agrees with the previous result. 

Centroid of an area. There are a large number of engineering 
problems in which we require to find the point in an area which 
would be the centre of gravity of a thin uniform flat sheet of the 
same contour as the area. An area has no weight, so that it is 
not strictly correct to speak of the centre of gravity of an area ; 
it is therefore called the centroid. Many people, however, use the 
term centre of gravity for both cases. 




Fig. 119. 

In the case of areas we may define the moment of an element 
of area about a line as the product of the element by its perpen- 
dicular distance from the line. 

Referring to Fig. 119, a is an element of area situated around 
a point P, then a xPN is the moment of the element about XX. 
If the whole area is divided up into elements and the moments of 

A. M. 14 



210 CENTRE OF GRAVITY AND CENTROID [ch. 

the elements are added together, the result is called the moment 
of the whole area. 
This is written 

Moment of area about XX = S {a . PN). 

From this point of view we may define the centroid as the 
point at which we can consider the whole area concentrated to 
give the same moment about any line. 

If ^ is the area and it is considered as concentrated about the 
centroid C, then A x dis equal to the moment of area about XX ; 

E{a.PN) 



d = 



(5). 



This is equivalent to the result we obtained in equations (3) 
and (4) for determining the position of the centre of gravity for 
solid bodies and we may take it that all the rules for finding the 
centre of gravity of a solid body can be applied to finding the 
centroid of an area. 



Centroid of a triangle. Let ABD, Fig. 120, represent a 
triangle and let HJ be a very narrow strip drawn parallel to 
the base. Since this strip is so 
narrow it may be considered as 
a rectangle and its centroid is 
therefore at its centre K. The 
whole triangle may be con- 
sidered divided up into strips, 
the centroid of each of which 
will be along the line AE which 
bisects the base at E and is 
called a median line. Therefore 
the centroid of the whole triangle 
must lie on AE. 

Similarly if we considered 
strips parallel to AD we should show that the centroid of the 
whole triangle must lie upon BF where F bisects AD. 

The centroid of the triangle must therefore be at the inter- 

EA 




section C of the median lines, and CE will be equal to 



3 



this 



is proved as follows. Join FE, then by a well-known geometrical 



xin] CENTRE OF GRAVITY AND CENTROID 211 



property of the triangle, FE will be parallel to AB and will be 

,. AB 
equal to —^ . 

Therefore the As ABC, CFE are similar; 

GE^EF _\ 
'''AC AB~2' 
AC 

2 ' 
AE 

3 • 

We get therefore the rule that ^'the centroid of a triangle is 
along a median line and is at a distance from the base equal to one- 
third of the height." 

Centre of gravity of a triangular pyramid. Let ABDE, 
Fig. 121, represent a triangular pyramid and let C^ be the centroid 



OP 



/. CE = 

CE = 





of the base. Join C-^A and consider a plane section FGH drawn 
parallel to the base BED. AC-^ cuts this section in C2, and it 
can be proved by an application of the principle of similar 
triangles that Og is also the centroid of the A FGH. Therefore 
the line AC^ passes through the centroids of all the plane sections 
drawn parallel to the base so that the centre of gravity of the 
whole body must lie upon the line AC^; similarly it must also 
lie upon EC^, so that the centre of gravity of the body is at the 
intersection G of AC^ and EC^. Now consider the section 
AEK of the pyramid through the edge AE and the point K; 
this is shown on the right-hand side of the figure re-drawn for 

14—2 



212 CENTRE OF GRAVITY AND CENTROID [ch. 
greater clearness. It contains the point G. Now consider the 

Zr~3 ^^ EK ~3 

[from previous proof for the triangle]. 

Therefore the As are similar and C-fi^ is parallel to AE and 

,. AE 
IS equal to k-. 

Next consider the As GCfi^ and GAE; their corresponding 
sides are parallel so that they also are similar; 

•• AG AE 3' 

:, GC^ = ^AG, 
or GC^^iAC^. 

We see therefore that ^'for a triangular pyramid the centre of 
gravity is on the line joining the apex to the centroid of the base and 
is ai a height from the base equal to one-fourth of the height of the 
pyramid** 

Extension to polygonal pyramid and cone. A polygonal 
pyramid (i.e. one whose base has more than three straight sides) 
may be divided up into a number of triangular pyramids of the 
same height the centre of gravity of each of which will be at 
one-fourth of the height of the base, so that the centre of gravity 
of the polygonal pjramid will also be at one-fourth of the height 
on a line joining the vertex to the centroid of the base. 

A cone may similarly be considered as divided up into an 
infinite number of triangular pyramids with sides radiating from 
the centre of the base so that in the cone also the centre of 
gravity is at a distance from the base equal to one-fourth of the 
height. 

Centroid of a trapezium. A trapezium is a four-sided figure 
with two sides parallel [some writers call it a "trapezoid"]. 
Referring to Fig. 122, if we considered narrow strips drawn 
parallel to the base, it is clear that the centroid of each must lie 
at the mid -point of each strip so that the centroid of the whole 
figure must lie somewhere upon the line FG joining the mid- 
points of the parallel sides. 

Draw BJ parallel to the side AE. We then have the figure 
divided up into a parallelogram ABJE and a triangle BJD, the 



xmj CENTRE OF GRAVITY AND CENTROIB 213 
cetitroids C^ and Cg of which will be at distances equal to ^ and 

■L 

^ from ED. We now require to find the distance d of the centroid 

G of the whole figure from ED. We have seen already that it 
must lie on FG and it must also lie on the line C^C^ joining the 
centroids of the two parts. 




Fig. 122. 

The area of the whole figure 

= Area of ABJE + Area of BJD 
y ib — a)h y / b — a\ 

= |(2a+6-a) 

= |(«+6) o,,.(l) 

= Half the height x Sum of the parallel sides. 
Now take moments about the base ED. 
Moment = M = Moment of ABJE + Moment of BJD 

= Area of ABJE x | + Area of BJD x | 

. h jb-a) h . h 
= ah.^ + 273 

'a b — a^ 

+ 



= -g (3a + 6 - a) 



^2 



{2a +b) {?:, 



214 CENTRE OF GRAVITY AND CENTROID [ch. 
But M = Area of whole figure x d 

«=|(a+6).(f; 

/. |(a+6).tZ = |(2a+6), 
, h/2a+b\ 

-lill^ (^)- 

It is interesting to note that if a = 0, d = s . ^^is being the 

case for the triangle ; whereas ifa = 6, d = k(i + o/'^s •2'^2' 

which is the result for the parallelogram. As a check therefore 
we note that the centroid of a trapezium is at a distance from 
the base somewhere between one-third and one-half of the height. 

Graphical construction. The following graphical construction 
is based upon the result of formula (3) and is very useful in many 
problems. 

Set out BK, Fig. 123, = 6 and EH = a and join across as 
shown. Then the intersection of HK and FG gives the centroid 
C required. 




Fig. 123. Graphical construction for centroid of trapezium. 

Graphical construction for centroid. The position of the 
centroid of any figure can be obtained by the following construc- 
tion which is a special case of the link and vector polygon 
construction (p. 28). 

Divide the area, Fig. 124, up into a number of small strips 
of equal breadth, parallel to the direction about which moments 
are taken, and draw the centre line of each of the said strips. Then 
if the strips are sufficiently small (we have only taken a few strips 
in the figure to avoid comphcation) the lengths of these centre 



xm] CENTRE OF GRAVITY AND CENTROID 215 

lines represent the areas of the separate strips. Now, on a vector 
line, to some scale, set out 1, 1 2,. . .6 7 to represent the area 
of each strip, and take a pole P at convenient distance = p from 
this vector line. Then anywhere across space draw and produce 
a line ah parallel to OP ; across space 1 draw ab parallel to PI ; 
across space 2, be parallel to P2, and so on until the point g is 
reached. Then draw the last link gh parallel to the last line 
P7 to meet ah in h. 




Fig. 124. Graphical construction for centroid. 

Then the centroid lies on the dotted line through h drawn 
parallel to the given direction. 

In most cases in practice we do not require the actual position 
of the centroid but only its distance from a line drawn in a given 
direction. In this case the above will suffice, the lines being drawn 
parallel to the given direction. If it does not, the construction 
should be repeated with the lines drawn at some convenient 
inclination — say at right angles — to the previous ones. 

It is not really essential to divide the area into strips of equal 
breadth ; any breadths may be taken, but in that case the areas 
of the strips must be set out on the vector line instead of the 
mid-ordinates. 

When the figure can be divided up into a number of figures 



216 CENTRE OF GRAVITY AND CENTROID [ch. 



the oentroids of each of which can be found by inspection, we 
proceed as in Fig. 126. 




Fig. 126. Graphical constniction for centroid. 
Draw lines parallel to the direction of the centroid line 
required through the separate oentroids c^, Cg, Cg and make the 
distances 1, 12, 2 3 on the vector line proportional to the 
separate areas and then proceed as before. 

Graphical constructions for any quadrilateral. The following 
graphical constructions for the centroid of an irregular quadri- 
lateral are useful. 





Fig. 127. Constmction for 
centroid of quadrilateral. 



Fig. 126. Construction for 
centroid of quadrilateral. 

First method. Let E be the point of intersection of the 
diagonals AG and BD, Fig. 126 ; from G set off CE' = AE and 
join DE' and BE', then the centroid of the quadrilateral will be 
the same as that of the A BE'D. Therefore bisect BE' and E'D 
in K, H and join DK, BH ; then their point of intersection 
gives the required centroid. 



XIII] CENTRE OF GRAVITY AND CENTROID 217 



Second method. Divide each of the sides into three equal 
parts and join across as indicated in Fig. 127. The resulting 
figure gives a parallelogram whose centre G is the centroid 
required. 

Numerical Examples. (1) Find the weight and centre of 
gravity of a cast iron body consisting of a cylinder 6 inches in 
diameter and 9 inches long^ with a cone of the same diameter and 
6 inches high standing on the top. Cast iron weighs '26 lb. 'per cu. in. 



The volume of the cylinder = 



TTd^h 77 X 62 X 9 



4 4 

= 8l7r; 
Weight of cylinder = SItt x -26 = 66 lbs. nearly. 

Trd^h TT X 62 X 6 



Volume of the cone = 



12 



12 
= IStt; 
/. Weight of cone = IStt x -26 = 14-7 lbs. ; 
.-. Total weight = 66 + 14-7 = 807 lbs. 

The centre of gravity of the cone 
and cylinder are at Gi and G^ respec- 
tively, Fig. 128. 

The centre of gravity G of the whole 
body will lie upon G1G2 and regarding 
the separate weights as acting at right 
angles to the plane of the paper we can 
take moments about any convenient 
point, say G'g* 

Then 
80-7 X GG^ = 14-7 x 6 + 66 x 0; 
14-7 X 6 



GG, 



80-7 



— 1*1 inches. 




Fig. 128. 



Therefore the centre of gravity is 
at 1*1 + 4*5 = 5*6 inches from the base 
of the cylinder. 

As an exercise the student should check this result by taking 
moments about G^, 



218 CENTRE OF GRAVrTY AND CENTROID [oh. 



(2) Find the position of the centroid of the cast iron beam 
section shown in Fig. 129. 

The centroid G is obviously upon 
the line of symmetry YY. To find 
its distance d from the base divide 
the section up into three rectangles 
as indicated, the area of each being 
regarded as acting at the centre. 

Then we have total area 

^ = 2xlJ + 7xl + 6xli 

= 19 sq. ins. 




Taking moments about the base 
we have 

Ad=^x 9-25 + 7 X 5 + 9 X -75 
= 27-75 + 35 + 6-75 

= 69-5 in. units ; 

.*. d = -^ = 3-66 inches. 

As an exercise the student should 
check this by the graphical con- 
struction shown in Fig. 125. 

(3) Find the position of the centroid of the angle section shown 
in Fig, 130. 

In this case we have a section which 
has no axis of symmetry and the cen- 
troid of which will lie outside the sec- 
tion. Divide up into two rectangles 
as shown. 

Total area = ^ = 4JxJ + 3xJ 
= 2-25 + 1-5 = 3-75. 

Take moments about ABy 

Ad^ = 1-5 x -25 4- 2-25 x 2-75 
= 6-562; 
6-662 




Fig. 130. 



^* ~ 3-75 



= 1-75 ins. 



XIII] CENTRE OF GRAVITY AND CENTROID 219 

Take moments about AD, 

Ady = 2-25 X -25 + 1-5 x 1-5 

= 2-81; 
ji 2-81 ^^ . 

(4) A circular disc 8 inches in diameter has cut out of it a 
circle of 3 inches diameter, leaving 2 inches on one side as indi- 
cated in Fig. 131. Find the centre of gravity of the resulting body. 

The area of the whole disc = — :: — . 

4 

TT X 3^ 

The area of the piece cut out = — j — . 

.*. The relative weights are 8^ and 
32, i.e. 64 and 9. 

.*. The relative weight of the re- 
mainder = 64 — 9 = 55. 

The line XX is a line of symmetry 
so that the centre of gravity G lies 
upon it. 

The centre of gravity of the whole 
disc is at (r^ and of the small circle at(r2. 




Take moments about (r. 



Fig. 131. 



Then 



55. GGi = miGji, 
55GG^ = 9 X -5, 

^^i = ~55~ 



= -082 in. 



Centroid of various figures. The positions of the centroid 
of the following figures are useful in calculations, but their proof 
is beyond our present scope. 





(b) 



Fig. 132. 



220 CENTRE OF GRAVITY AND CENTROID [ch. 

Parabola. Fig. 132 (a). The area of the interior segment 

XYZ = \bH and of the exterior segment XZU = ^. The 
o o 

centroids G^, G^ are as indicated. 

Semicircular arc. The centre of gravity of a rod bent to a 

2r 

semicircle will be at Gy Fig. 132 (6), where rf = — . 

IT 

Semicircular area. The centroid G is given hyd = ^— . 

Experiments upon centre of gravity and centroid. Centre of gravity of a 
plate by suspension. The centre of gravity of a plate or lamina can be found 
by hanging it up by one point A, Fig. 133, and drawing a line on the plate 
continuing the direction of the string, etc. The line then passes through the 

////////////// 




Fig. 133. 

centre of gravity. The plate is then suspended from some other point B, 
preferably in about the relative position shown, and another similar line drawn. 
The intersection gives the centre of gravity Q. 

Centre of gravity of a walking stick. An interesting but very simple 
experiment can be performed with a walking stick as follows. Hold the stick 
horizontally with one finger near eftch end m indicated in Fig. 134 Then 



CH 



MEBO-'iKil 




T> 



tp- 



Fig. 134. 



xin] CENTRE OF GRAVITY AND CENTROID 221 



move the fingers A, B towards each other fairly slowly without jerking and the 
fingers will meet at the centre of gravity. 

After the student has finished reading this book, he should try to think out 
why this gives the centre of gravity. 

Kinds of Equilibrium. Directly the line of pressure of a 
body falls outside the base, a moment acts which will make 
the body topple over; but if the line of pressure falls inside 
the base, the moment acting tends to maintain the body in 
equilibrium*. 

It is common to speak of the equilibrium of a body as being 
one of three kinds, Fig. 135 : 

Stable equilibrium, in which the body tends to return to its 
original position of equilibrium when given a slight displacement. 

Unstable equilibrium, in which the body tends to lose its 
equilibrium when given a slight displacement. 

Neutral equilibrium, in which the body neither returns to its 
original position nor loses its equilibrium. 




77/777777777777777777 

Stable 
equilibrium 




Unstable 
equilibrium 

Fig. 136. 



^77777777777^ 

Neutral 
equilibrium 



SUMMARY OF CHAPTER XIII. 

The centre of gravity of a body is the point at which the resultant 
weight of the whole body may be considered to act. 

y,_2WiXi _Siim of moments of separate weights 
2Wi Sum of separate weights 

The centre of gravity is the point about which a body will balance. 
The centre of gravity of a body lies upon a section of synunetry. 

The centroid of an area is the point at which the whole area 
may be considered concentrated to give the same moment about 

♦ Cf. p. 22. The student should prove as an exercise that if the over- 
turning moment exceeds the stability moment, the Une of pressure will fall 
outside the base. 



222 CENTRE OF GRAVITY AND CENTROID [ch. 

any line. It is often called the centre of gravity, but strictly this 
is not correct because an area has no weight. 

The centroid of a triangle is along a median line at a distance from 
the base equal to one-third of the height. 

The centre of gravity of a pyramid or cone is on the Une joining 
the apex to the centroid of the base and is at a height from the base 
equal to J the height of the pyramid. 

For a trapeziimi of height h^ the distance of the centroid from 

the base b is o(H"~T/i]» ^ being the side parallel to the base. 

EXERCISES. Xin. 

1. Find the position of the centroid of an isosceles triangle 
4 inches base and 6 inches high. 

2. Find the position of the centre of gravity of a cone 10 inches 
high and 8 inches diameter at the base. 

3. A uniform rod of 5 lbs. is weighted with weights of 1 and 
2 lbs. at the ends. Find the point about which it will balance. 

4. Find the position of the centre of gravity of a square, length 
of side 2 ft., from which is cut out a circle of 1 in. diameter touching 
one of the sides at the centre. 

Find the centre of gravity of the following : 

5. A rod of length 2 feet weighing 2 lbs. to the end of which 
is fixed a spherical ball weighing 10 lbs. and 4 inches in diameter. 

6. A T-shaped figure; the stem being 3 feet x 3 inches wide 
and the top 12 inches wide and 4 inches deep. 

7. A balance weight having the form of a circular quadrant 
of radius R. 

8. A trapezoidal wall 30 ft. high has a vertical back and sloping 
front face. The base is 10 ft. and top 7 ft. wide. What force 
must be applied horizontally at a point at 20 feet from the top to 
overturn it? Take width of wall = 1 ft. and weight of masonry 
130 lbs. per cu. ft. 

9. A figure is made up of a square upon which stands an isosceles 
triangle. Find the relation between the height and base of the 
triangle in order that the centroid of the whole figure may be in 
the common base. 

10. Find the position of the centroid of a channel section of 
base 10 inches, sides 3 inches and thickness of metal | in. 

11. Find the centre of gravity of the given figure. (See Fig. 
XIII a.) 

12. Find the centre of gravity of an angle iron 4^x3"x^'. 



XIII] CENTRE OF GRAVITY AND CENTROID 223 



13. Find the distance of the centre of gravity of the trapezium 
ABCD from CD (Fig. XIII 6). 

14. A rod 5 ft. long has a weight of 2 lbs. at one end and 3 lbs. 
at the other, also a weight of 6 lbs. at centre. Neglecting the weight 
of the rod, find the point about which it will balance. 

15. A ship with equipment weighs 6000 tons. How far will 
its centre of gravity move if a gun weighing 30 tons is moved 
20 feet across the deck? 

•i 4 ■> 



10 



4 



-h 



A 

!i1 



\V4 



X< f- ^Y 

Fig. Xm a. 





C< 12" ^D 

Fig. XIII b. Fig. XIII c. 

16. The bending moment of a beam of span I is made up of a 

triangle of height — ^^ *^® centre and a parabola of height ^ 

extending from the right-hand end to the centre (Fig. XIII c). Taking 
the area of a parabola = f base x height find the position of the 
centroid of the diagram from one end. 

17. A solid cone 2 ft. high on a circular base has | of its volume 
removed, being cut by a plane parallel to its base. Find the position 
of the centre of gravity of the remainder. 

18. A circular disc 6 feet in diameter has a circular hole 6 inches 
in diameter cut out from it, the centre of the hole being 2 feet from 
one edge of the disc. How far will the centre of gravity be from 
the nearest edge? 



CHAPTER XIV 



FRICTION AND LUBRICATION 



m7T77Trm7Tm\7T777mTmT77T 



We have explained already that a resistive force called 
friction is the principal cause of the loss of energy in machines 
and also that in some cases, such as in road traction, this frictional 
force is of great use. We wiU now consider the subject in greater 
detail and would ask the student to try to grasp fully each 
point as he proceeds, because this is a branch of the subject which 
is not always understood very clearly by students. 

Static and kinetic friction. Let A and B, Fig. 136, be two 
bodies pressed together with a normal pressure P. Then since 
this force P has no component 
at right angles to itself (i.e. in a 
horizontal direction in the figure) 
there should be no force required 
to cause a sliding motion of A 
upon B. But actually there is 
a force tending to resist this 
sliding motion, and this resistive 
force is called the /orce of friction. 

Now as the force F is slowly increased the resistive force/ 
increases also, but we soon reach the condition when the sHding 
motion will commence because the force / is not capable of 
exceeding a certain value called the limit of static friction. The 
word static is used because the bodies are relatively stationary, 
and some writers have used the ingenious term stiction for it. 
As F gets stiU larger motion takes place and a frictional force 
/ still comes into play, but it will not be quite equal to the limit 
of static friction and will depend to some extent upon the speed 
of sliding. The frictional force is then called kinetic friction or 
friction of motion. 



Fig. 136. 



B 

P 
Friction. 



CH. XIV] FRICTION AND LUBRICATION 



225 



f 
Coefficient of Friction. The quantity ^ is called the coefficient 

of friction and is generally given the letter {x. 



'-i 



«(1) 



or f=fxP (2). 

If therefore we are given the value of the coefficient of friction 
for the materials and conditions under consideration we can at 
once find the value of the friction force when the pressure between 
the surfaces is given. 

With regard to static friction, the values of /x which are 
tabulated in the various books are those for the limiting friction. 
It should be remembered that the frictional force only becomes 
equal to [jlP at the moment when slipping is about to occur; 
until this condition is reached, the friction force / will be equal 
to the force F because if there is no relative motion between the 
two bodies the forces must be in equilibrium. 

It has been found by experiment that for two given materials 
the coefficient of friction with dry surfaces is practically constant 
for various pressures ; it is, however, a little smaller for a large 
pressure acting upon a small area than for a small pressure acting 
upon a large area. 

Limiting reaction with friction; angle of friction. We have 
already seen that with smooth surfaces the reaction is always 
normal, i.e. at right angles to the 
surface*. With rough surfaces 
the reaction will be inclined to 
the normal in such a manner as 
to tend to oppose the motion. 
In the limiting condition when 
motion is just about to take 
place, the reaction reaches its 
limiting position and the angle 
at which it is inclined to the 
normal is called the angle of 
friction. 

We will explain this more 
fully with reference to a diagram, Fig. 137. 

♦ p. 59. 




Kg. 137. Angle of Friction. 



A.M. 



15 



226 



FRICTION AND LUBRICATION 



[oh. 



The body A rests upon the body B and a force P acts in a 
direction at right angles to their surface of contact; a force 
F acts parallel to the surface and the body ^ is in equilibrium. 
Between the two surfaces there acts the friction force / indicated 
by the small arrows in the figure and a reaction pressure P which, 
in accordance with Newton's third law, is equal and opposite to 
the force P. 

The reaction pressure P and the friction force / have a 
resultant reaction R which is inclined at an angle ^ to the normal ; 
this angle is called the angle of friction. 

We may therefore define the angle of friction as follows : 

The angle of friction is the angle with the normal which is made 
by the resultant reaction between two surfaces when slipping is about 
to take place. 

Referring again to the figure we note that abc is a triangle of 
forces and that 



be 
ba 



= tan^ 



-^L-f^I-. 



fX, 



Therefore we obtain the rule that : 

The tangent of the angle oj jriction is equal to the coefficient of 
friction. 

Average values of ft. The following values of /x and </> may 
be taken as average values for dry surfaces. 



y 



Surfaces 


Coefficient of 
friction n 


Angle of 
friction 


Oak on oak (along grain) 

„ „ (across grain) 

Wrought iron on wrought iron .... 
Cast iron on cast iron 


•48 
•34 
•14 
•15 
•49 
•16 
•64 
•33 


25° 38' 

18° 47' 

8° 0' 

8° 30' 

26° 6' 

9° 6' 

32° 30' 

18° 15' 


Cast iron on oak (parallel to grain) 
Brass on wrought iron 


Common brick on common brick . . 
Masonry on moist clay 





Numerical Examples. (1) A block weighing 30 lbs. rests 
upon a rough plate whose coefficient of friction is -2. Find the 
least force acting horizontally which will move it. 



xiv] 



FRICTION AND LUBRICATION 



227 



In this case P = 30 lbs. and fi = -2, 
.*. Limiting friction = fxP 

= -2 X 30 
= 6 lbs. 
.*. Any force exceeding 6 lbs. will move the block. 
(2) In the above case, what is the least inclined force which will 
move the block and what will be its direction ? 

We can solve this problem by considering the triangle of forces. 
Referring to Fig. 138, the three forces acting upon the block are 





Fig. 138. Friction with inclined force. 

the weight of 30 lbs., the resultant reaction R acting at an angle 
<j) to the normal in a direction tending to oppose the motion and 
the tractive force jP, whose direction we do not yet know. 

We are given that /Lt = tan <f) = -2, and from trigonometrical 
tables we find that the value of ^ is about 11° 20'. 

To draw the triangle of forces we set down ab to represent 30 lbs. 
to a convenient scale and then draw bd at an angle ^ to it. Then 
if F were horizontal we should draw ad horizontally to fix the 
point d, and if F were in any other given direction we should 
draw from a a parallel to it. It will be clear that the force F is 
least when the distance from a to ic^ is the least possible, i.e. when 
ac is drawn at right angles to bd as shown. 

If the triangle be drawn accurately to scale and ac be measured 
it will be found to be about 5*88 lbs. 

By calculation we should say 

F ac . , 

/. i?' = 30sinll°20' 

= 5-88 lbs, approx. 

15—2 



228 FRICTION AND LUBRICATION [ch. 

To determine the direction of F draw ce horizontally, then 
L ace = 90° — l cae = <^ [from Aa6c]. 

We see therefore that the best direction in which to pull a 
body along a rough surface is at an angle to the surface equal to the 
angle of friction. 

Rolling Friction. It is a fact of universal experience that 
it is easier to push an article provided with wheels than to push 
one without. This is often explained by saying that rolling 
friction is less than sliding friction, but such explanation does 
not get us much further. The exact nature of so-called rolling 
friction is not understood, but in a pure rolling motion there is no 
sliding motion at all and as frictional forces are solely brought 
into play by sliding there is no friction in pure rolling; on this 




^mffmimirm -mTJTrJlTrwtflTrnTTTr. 

Fig. 139. Rolling Friction. 

argument we should expect that with a very hard bed and roller 
the frictional resistance would be practically nothing. In most 
cases in practice, however, the roller or wheel sinks somewhat into 
the bed as indicated in Fig. 139 (6) and so the pure rolling action 
is stopped and some slipping occurs thus introducing friction. 
There is also a resistance due to the fact that the wheel is in a 
sense always going slightly uphill on account of the hump formed 
in the front of the depression. The harder the surface, the less 
wlQ be the rolling resistance. Cyclists who have ridden upon 
very soft sandy roads and then come on to hard tarred roads will 
already have appreciated the truth of the above rule. 

We should also expect the resistance to rolling to be less for 
wheels of large diameter than for small ones, because the smaller 
one will make a deeper depression than the larger one. Experi- 
ments show that this is true and that the following two rules are 
approximately true also : 

(1) The rolling resistance is proportional to the load. 

(2) The rolling resistance is inversely proportional to the 
diameter of the roller. 



XIV] FRICTION AND LUBRICATION 229 

Action of wheels in assisting traction. In the case of a vehicle 
such as a cart it might be argued that even if you have a wheel 
you still have a sliding action at the axle which induces a friction 
there instead of at the road. Against this we can point out that 
the axle is iron running upon wood, brass or iron and that it is 
easy to lubricate the axle; but this is not a complete answer. 
The important poiat is that we have made the friction work at 
a very small radius and the wheel gives a leverage over this 
which makes the cart much easier to pull. Referring to Fig. 139 {a) 
we see that the friction / acts at the axle circumference and the 
tractive force F causes an equal resistance at the ground and this 
force F has a large leverage over the friction force /. 

Another way of looking at it is that in one revolution of the 
wheel the work done against the friction is / x 2Trr and the work 
done by the traction is ^ x 27tR. If the tractive force is just 
sufi&cient to move the cart horizontally, these two amounts of 
work will be the same; 

/. / X 27Tr = jP X 27tB, 

This is the same result as we should obtain by a consideration 
of leverage. 

It is important to remember that rolUng resistance is small 
only if the road is hard. It is easier to puU a flat bottomed box 
along very soft sand than to pull the same box mounted on wheels. 
It is an interesting fact that about 150 years ago wheeled carts 
were practically imknown in the agricultural parts of Scotland 
(see for instance Smiles' Lives of the Engineers) ; this was doubt- 
less in part due to the fact that the roads were so bad that 
practically all the advantage of wheels was lost. 

Inclined plane and screw with friction. We have considered 
already (p. 58) the case of the inclined plane (of which the 
screw is a modification) in which frictional resistance was 
neglected. We will now consider the case in which friction has 
to be considered. 

Case 1. Force parallel to plane. m^^ 

(a) Body moving upwards. In this case the three forces 
acting upon the body are the force F, Fig. 140, parallel to the 
plane, the weight W vertically downward and the reaction E 



230 



FRICTION AND LUBRICATION 



[oh. 



which will be inclined at the angle of friction <f) to the normal 
to the plane (shown in dotted lines). 

We have already indicated (Fig. 33) that this normal is at an 
angle 6 with the vertical, so that the reaction R makes an angle 
{6 + cf)) with the vertical. We are thus able to draw the triangle 
of forces abc^ and the force F can either be found graphically or 




(80-?) „\A 



Figs. 140, 141. Inclined plane with Friction Force parallel to plane. 



by calculation by means of a trigonometrical solution of the 
Aa6c. Etud'mts who have gone sufficiently far with their 
trigoDO^«3try 'rill understand the following solution; 

F ac _ sm{d + (f>) 

W~ ab~ sin (90-^) 

8in(^+^)^ 



COS0 ' 

, j^_W8m{e+cl>) 

e e X" — - . 

C08(j) 

^ = ^ = sin (9Q - ^) 
W ab sin (90 - (ji) 

cos d 



/. R = 



cos (j) ' 

W cos d 

COB(f) 



(1). 



(2). 



xiv] 



FRICTION AND LUBRICATION 



231 



(b) Body moving downwards. In this case the reaction still 
acts at an angle (f) to the normal, but on the opposite side of it, 
so that it makes an angle (^ — </>) with the vertical. 

The triangle of forces abc is then as in Fig. 141 and by similar 

reasoning we get 

W sin {9 - cf)) 



F = 



E== 



C08<f> 

Wcosd 

COS(f) 



(3), 



(4). 



It will be noted that R is the same in both cases. 



Case 2. Force horizontal. 

(a) Body moving upwards. In this case we have in Fig. 142 
the triangle of forces abc, B being as before at an angle (j> to the 
normal. 





Fig. 142. Friction on inclined plane. Force horisocta]. 

We then have from the triangle abc 

F ac , ,n IV 

/. F=W tan (B + (j>) 

R be iQ , i\ 

w=:;;A = cosec(^+^); 



(OSes 



,,(6), 



W ab 

:. B = W coBec {d + (l>) «.,.>-.-., ,(6). 

(6) Body moving downwards. In this ca,se t70 shall have 
as in Case 1 the reaction B still at an angle ^ to ^ho normal but 
it will be on the opposite side, so that it will ba ioclined at an 
angle {6 — <j)) to the vertical and we shall hi\^-f^ hy a similar 
consideration of the triangle of forces 

#=irtan(^-^) (7), 

B=W cosec (B-<l>) (8). 



232 FRICTION AND LUBRICATION [ch. 

Numerical Examples. (1) A weigJd of 20 lbs. rests upon an 
inclined plane whose base is 4 feet long and whose height is 3 feet 
and is just prevented from moving downwards by a force of 8 lbs. 
acting horizontally. Find the coefficient of friction. 

In this case the body is about to move downwards and 
F = 8 lbs. and TT = 20 lbs. 

Therefore by equation (7) 

8 = 20 tan {d-<f>); 

:. tan(6l-^) = ^=-4. 

A table of tangents shows us that tan 21 '8° is approx. '4, so 

that we have {d - ^) = 21-8°. 

Now tan ^ = f = '75 and from tables we have 6 = 36*9° 

about ; 

/. ^ = 36-9 - 21-8 = 14-7° about, 

.*. fx = tan 15* P 

= -27 about . 

(2)* A block weighing 60 Ihs. is on the point of motion down 
a rough inclined board when supported by a force of 24 lbs. acting 
parallel to the board and just begins to move up when acted upon 
by a force of 36 Ihs. also parallel to the board. What is the coefficient 
of friction ? 

In this case we have, when about to move down, by 
equation (3) 

60sm_(^_-^ 

COS^ 
• ,r\ IV 2C0S^ 

I.e. 8m(^-^)= — ^— ^ ...(9). 

When about to move up we have by equation (1) 



^^ 60sin(^ + (/>) 

COS^ 



- iQ , i\ 3cos^ 



Now sin {9-(f)) = sin 6co8<f> — cos 6 sin cf) 

and sin (9 -\- cf)) = sin 6 cos <f> + cos 6 sin (/> 



....(10). 
....(11). 



• Students who do not possess fair knowledge of trigonometry will not be 
able to follow this example. 



xivj FRICTION AND LUBRICATION 233 

.*. putting these values in (9) and (10) and adding we get 
2 sin 6 cos (f> = cos cj) ; 
.*. sin ^ = J, 
or d= 30° ; 

.*. sin (6 + (f)) = sin 30 cos cj) + cos 30 sin ^ 
_ cos sin ^ VS 

, ,,^, 3cos(^ cos^ sin<^\/3 

.-. by (10) — 1,- = -^ + 



1.0. 



5 2 ' 2 

sm/ 



iiiKhVs , /3 1\ COS0 

sin <^ _ 1 
COS 9!) ~ 5 V3 * 

i.e. ft = *115. 

Angle of Repose. The largest angle of an inclined plane 
upon which a body can rest without sliding down is called the 
angle of repose, and we can show in the following manner that 
the angle of repose is equal to the angle of friction. 

When a body is just about to slide down, the force acting 
either parallel to the plane or horizontally is zero, so that by 
equation (7) we have 

= TT tan {6 - ^). 

Since W is not zero, tan {6 — (j)) must be zero, i.e. (0 — <^) = or 

The angle of repose is of importance in considering the 
stability of walls supporting banks of earth. 

The efficiency of a screw. We have shown on p. 63 that 
a screw is really a special case of an inclined plane with a horizontal 
force, and when friction was neglected we had the relation 

^= TFtan^ (12). 

When friction is considered we get the following treatment : 
(a) Scremng in. When screwing in, the load is moving up 
the plane, so that equation (5) is the one to use. 
We have therefore 

i' = IF tan (^ + ^) (5). 



234 FRICTION AND LUBRICATION [ch. 

Now the efficiency (7) of a machine may be determined by 

the relation 

_ Ideal effort 

^ Actual effort 
Tf tan^_ 
^) 

(13). 



Wtdin(d+cj>) 
tan 9 



tsin{e+cf>)* 

It can be shown that this efficiency is the maximum possible 

when 6 = 45^ — ^ , but the proof of this is beyond our present 

standard. 

(6) Screwing out. When screwing out, the load is moving 
down the plane, so that equation (7) is relevant; 

/. F=W tan {d - <j)). 
In this case F is the horizontal force that the weight W will 
move downwards, the effort and resistance being reversed. 

F 



Actual effort = 
Ideal effort = 



tan {d-<f))* 
F 



tan^* 

_ Ideal effort 
^ Actual effort 

^tan(e-i) 

tan d 
This is found to be a maximum when 

6'=45°+|. 

It will be noticed that if equation (14) = 0, it means that the 
screw will not run backwards unless it is helped round. The 
screw is then called self-locking and this in many machines is 
a useful feature but it means that the efficiency of the machine 
is sacrificed to it. 

This occurs when 6 = cf) ot if the angle of a screw is less than 
the angle of friction the screw will not run backwards, i.e. the nut 
will not drive the screw. 

A 4. /a . ^>. tan ^ + tan d> 
^ ^' l-tan<?tan0 



xiv] 



FRICTION AND LUBRICATION 



235 



Ladder resting against a wall. If a ladder AB, Fig. 143, 
rest against a wall at B and against the ground at A, frictional 
forces are induced at A and B preventing the ladder from sliding 
down the wall. If slipping is about to take place, the reactions 
jB^ and Bb will be inclined as shown at angles ^^ and (f)^ to the 
normal, ^^ being the angle of friction for the ladder on the ground 
and ^2 ^or the ladder along the wall. 




Fig. 143. Ladder resting against a wall. 

Now we have already proved (p. 26) that if three forces are 
in equilibrium they must, if not parallel, pass through a pomt. 
As therefore B^ and Bb meet at (7, the resultant weight W, of the 
ladder and of a man standing upon it, must also act through C. 

It must always be remembered that in these friction problems 
it is only when slipping is just about to take place that the 
reactions are inclined at the angle of friction. In other cases, 
as for instance when the resultant weight of the ladder comes 
below the point D, the reactions will be less inclined and their 
actual values cannot always be determined. In the ladder 
problem, all that we know is that the reactions must intersect 
on the vertical line through the resultant weight and that neither 
reaction can be inclined to the normal at an angle greater than 



236 



FRICTION AND LUBRICATION 



[CH. 



the angle of friction but there are a very large number of reactions 
possible which will satisfy these conditions. Problems of this 
kind in which the exact result cannot be found are called 
' ' statically indeterminate.' * 

Numerical Examples. (1) A wheel rotates upon an axle 
3 inches in diameter and makes 90 revolutions per minute. If the 
load on the wheel is J ton and the coefficient of friction for the 
lubricated axle is '02, how much work per minute is absorbed in 
friction ? 

Fig. 144 shows the axle, the 
weight being regarded as acting at 
the bottom. 

In this case the load 

« Tr= 2^0 = 560 lbs.; 
.*. Friction force = fiW 

= -02 X 560= 11-2 lbs. 
Now the axle is constantly ro- 
tating in opposition to this friction. 
.*. Distance moved per minute 

= ttDN Fig. 144. 

77 X 3 X 90 -, 

= I2 **• 

= 70-7 ft. 
/. Work done against friction per minute 
= Force x Distance moved 
= 11-2 X 70-7 
= 792 ft.-lbs. 

We have taken the weight as acting at the bottom in this 
case because this is approximately true ; strictly the weight will 
act a little to the right of the bottom, suflSciently far away for 
the resultant of the normal reaction and the friction force to be 
exactly equal and opposite to the weight. 

(2) Find the efficiency of a screw 2 J inches in diameter in 
which there are four threads to the inch and the coefficient of friction 
is -04. 

In this case p = \ inch and the circumference = 2-57r; 




tan d = \ ^ 2o7T = T^r- = '032 approx. 
IOtt 



xiv] 



FRICTION AND LUBRICATION 



237 



Now by equation (13) 

_ tan^ 
'^ ^ tan {9 + <t>) 



__ tan ^ (1 — tan 6 . tan<^) 
~ tan 6 + tan cf) 



•032 (1 - -0013) 
•032 + 04 



= •44; 



;•. Efficiency of screw = -44 or 44 % . 

(3) A cylinder weighing 6 lbs. is 2 inches in diameter and 
8 inches long. It is placed on a board which is slowly tilted wp. 
If the coefficient of friction between the board and the cylinder is '2, 
will the cylinder start sliding before it topples over ? 

We have already seen (p. 221) that if the line of action of 
the weight of a body falls outside the base it will topple over 
unless held down by some ex- 
ternal means ; we have also learnt 
that the body will start sliding 
when the slope is equal to the 
angle of friction. In Fig. 145 
we have shown the board at this 
slope (ii = tan <f> = ^= -2); we 
therefore require to find whether 
in this position the line of action of the weight of the cylinder 
falls outside the base. This can be done by drawing carefully 
to scale and then drawing a vertical through the centre of the 
cylinder. It wiU be found to come just inside so that the cylinder 
will sUde before it topples over. 

We can obtain this result by calculation as follows : the angle 
between a6 and ac will also be ^ ; 

be 




tan^ 



ab 



or 



6c = 4 X 



, but ab =4: inches, 

^~4 ' 

•2 = -8 inch. 



As the cylinder is 2 inches in diameter the distance from b to 
the edge of the base will be 1 inch so that c falls inside the base. 

Lubrication. The purpose of lubrication is to reduce friction 
and so minimise the energy which is absorbed by the friction. 
This is effected by imprisoning a film of oil between the two 
surfaces so that the friction between the surfaces is replaced by 



238 



FRICTION AND LUBRICATION 



[OH. 



a friction between the fluid and the surfaces and this is less than 
the friction between the dry surfaces. 

In the choice of a lubricant it should be remembered that 
the condition under which it is to be used should be considered. 
It should have sufficient viscosity to prevent its being squeezed 
right out of the bearing and if the part lubricated is likely to be 
hot in working the lubricant should be such that its lubricating 
properties are not destroyed at the higher temperature. In 
designing lubricating devices care should be taken that the 
lubricant is not introduced at the point where the pressure is 
greatest; otherwise little will find its way to the bearing. The 
friction of lubricated bearings is really a subject requiring 
separate attention and it is rather beyond our present stage. 

Experiments upon Friction. The following experiments can be made 
with very simple apparatus. 

(1) Determination of the coefficient of friction by tilting. Hinge a board Q 
at one end to a board A (Fig. 146), and at eaoh side of the opposite end of the 




Fig. 146. Determination of coefficient of friction by tilting. 

latter set up two slotted uprights B. Between the uprights fix a bolt D 
provided with a fly-nut by means of which it can be fixed in any position, 
upon which the board can rest. On the edge of the upright iS fix a scale 
E. The block F whose angle of friction with the board C is required is placed 
upon the board C which is slowly tilted upwards until the block begins to 
slide. 

The height h at which sliding commences is noted and then we have, as 
shown on p. 226, 

/»=tan&=^ . 

By choosing I a convenient round number of inches, the scale E can easily 
be graduated to read oflE values of fx. direct. To make an instructive experi- 
ment blocks F of dilVerent weights and areas of contact for the same material 



XIV] 



FRICTION AND LUBRICATION 



239 



may be taken so that the student can discover for himself what effect the 
pressure has upon the coefficient. 

(2) Determination of the coefficient of friction by weights. Fix a smoothly 
running pulley B (Fig. 147) in the end of a board A and connect a thin string to 
a block D, the string passing over the pulley and having a scale pan attached to 
its end. A weight is then placed on the block, the combined weight including 
that of the block being W ; small weights are then placed carefully in the scale 
pan until the block begins to slide. Then if / is the sum of the added weights 
and the weight of the scale pan, the limiting coefficient of static friction will 
be given by 

/ 



777777777777 
/ 



ZJ: 



/ 



777777777777777777^7^^7777777777777777;^ 



^ 



w 



B 



A 



Fig. 147. Determination of coefficient of friction by weights. 

The experiment may be extended by varying the weight W and by placing 
surfaces of various kinds upon the board A and also by turning the block D 
round to vary the direction of the grain. 

Experiments may also be made to find the coefficient of kinetic friction 
by loading the scale pan until when the block is given a start it will continue 
to move uniformly. 

(3) Experiments upon rolling friction. The same apparatus may be employed 
for experiments upon rolling friction by replacing the block D by a small model 
wheeled truck. The friction of the axles will of course come into play but 
the effect of various surfaces upon the friction for the same truck may be 
investigated by laying various surfaces upon the board. 



SUMMARY OF CHAPTER XIV. 

Friction is the force between two surfaces which tends to prevent 
them from moving relatively to each other. 

When motion is already taking place the force is called kinetic 
friction, but when the bodies are stationary the friction force when 
sliding is just about to take place is called the limit of static friction, 

fi= tan (f). 

Friction is reduced by replacing a sliding motion by a rolling 
motion. 



240 FRICTION AND LUBRICATION [ch. 

To deal with friction on an inclined plane treat the reaction as 
acting at an angle ^ to the normal to the surface on the side which 
will tend to oppose motion. 

The angle of repose is equal to the angle of friction 

Efficiency of screw = - — r^^ — — . 
•^ tan(^ + ^) 



EXERCISES. XIV. 

!• A weight of 5 cwt. resting on a horizontal plane requires 
a horizontal force of 100 lbs. to move it against friction. What in 
that case is the value of the coefficient of friction? 

2. A body weighing 40 lbs. rests on a rough horizontal plane 
whose coefficient of friction = 0*25. Find the least horizontal force 
which will move the body. 

3. A locomotive weighs 65 tons of which 0'48 of the whole 
rests on the driving-wheels. What must the coefficient of friction 
be between the driving-wheels and the rails so that the engine 
may draw a train of total weight 200 tons at 60 miles an hour up 
an incline 1 in 300 ? Resistance = 45 lbs. per ton. 

4. A horse drags a load of 35 cwt. up an incline of 1 in 20. 
The resistance on the level is 100 lbs. per ton. Find the pull on the 
traces when they are (a) horizontal; (6) parallel with the incline; 
(c) in the position of the least pull. 

5. If the angle of friction is 10°, find the magnitude and direction 
of the least force which will push a load of 20 tons up a plane inclined 
at 20° to the horizon. 

6. A bicycle and rider weighing together 180 lbs. are going along 
the level at 10 miles an hour. If the brake be applied at the top of 
the front wheel (30" diam.) and is the only resistance acting, how 
far will the bicycle travel before stopping if the pressure of the 
brake is 20 lbs. and [i = 0-5? 

7. Prove that a train going 60 miles an hour can be brought 
to rest in 313 yds. (about) by the brakes supposing them to press 
on wheels with f weight of the train and /* = O-IS in addition to a 
passive resistance of 20 Ibs.-wt. per ton on the level. 

8. A wheel 12 ft. in diameter, rotating at the rate of 1 rev. in 
2 seconds, is acted upon by a brake which applies normal pressures 
of 1 cwt. each at opposite ends of a diameter. If /u = 0-6, find the 
H.p. absorbed. 



XIV] FRICTION AND LUBRICATION 241 

9. If the coefficient of friction be f , find the least depth from 
back to front of a drawer 2 ft. wide, which can be drawn out by 
a direct pull on a handle 6 ins. to the right or left of the middle 
of the front. 

10. A ship weighing 2000 tons is launched. Find what slope 
of the ways is necessary for uniform motion when once started. 
Also what should be the area of the bearing surface so that the 
pressure shall not exceed 2^ tons per sq, ft. and so force out the tallow, 
fi = 0-14. 

11. In a screw-jack the pitch of the square-threaded screw 
is 0*5 in. and the mean diamieter is 2 ins. The force exerted on 
the bar used in turning the screw is applied at a radius of 21 ins. 
Find this force if a load of 3 tons is being raised. Taking /n = 0*2, 
what is the efficiency of this machine? 

12. A uniform ladder 70 ft. long is equally inclined to a vertical 
wall and horizontal ground, both of which are rough. The weight 
of a man and his burden ascending the ladder is 2 cwt. and the weight 
of the ladder is 4 cwt. How far up may he ascend before the ladder 
begins to slip if /x = ^ for the ground and ^ for the wall ? 



A. M. 16 



CHAPTER XV 



MOTION m A CURVED PATH 

The Hodograph. We have considered up to the present 
only the cases in which the motion of a body takes place in a 
straight line. If a body moves in a curved path, its motion may 
in many cases be considered most conveniently by means of 
the Hodograph which is defined as follows: 

Let Pq, Pj ... P4, etc., Fig. 148, represent successive points 
upon the curved path of a body and let PqO, P^l, etc. be the 
tangents to the curve at the various points. 




Curved Path . R 



Fig. 148. The Hodograph, 

Taking a pole X draw a vector XO parallel to the tangent 
to represent the velocity Vq of the body at the point Pq in 
magnitude and direction to some convenient scale; then draw 
XI parallel to P^l to represent the velocity v^ at P^ to the same 
scale and so on. Then the curve obtained by joining the points 
0, 1, 2 ... 4, etc. is called the velocity hodograph for the motion. 

Now consider the question of acceleration. Acceleration is 



OH. XV] 



MOTION m A CURVED PATH 



243 



defined as the rate of change of velocity, and the change in 

velocity may consist of a change of direction as well as one of 

magnitude. In the case under consideration for instance the 

velocity between the points Pj, Pg changes from XI to X2, the 

change in velocity being represented by the vector difference 

1, 2. If the distance P1P2 is very short and the time taken in 

traversing it is Bt, we have 

Change in velocity 



Acceleration = 



I.e. 



a = 



Time taken 

hi 
8t 



..(1). 



This means that the acceleration of the body between the 
points Pj , P2 is equal to the velocity with which the corresponding 
point in the hodograph moves across the corresponding period. 

This gives us the rule that ^'the velocity in the hodograph is equal 
to the acceleration in the curved path.'' The acceleration at any 
point will also be in the direction of the tangent to the hodograph 
at the point. 

If therefore we consider the velocity hodograph as a curved 
path and repeat the construction, the new curve will give accelera- 
tions and may be called the acceleration hodograph. 

Uniform motion in a circle; angular velocity. Suppose that 
a point moves with a velocity t; in a circle of radius r, and that 
in a time t the point moves through an arc AB, Fig. 149, sub- 
tending an angle 6 at the centre of the circle. 

Then the angle turned through 
in a unit time is called the angular 
velocity and is given the letter o). 

Then since 

arc = angle (in radians) x radius 
we have AB = rdy 

and if t is the time taken from A to B, 
AB = vt; 
.-. rd = vt 



or 



V = 



rd 




but J = angular velocity = co ; 



car 



....(2). 
16—2 



244 



MOTION IN A CURVED PATH 



[oh. 



In practice angular velocity is not expressed in radians per 
minute or per second, but in revolutions per minute or per second. 

Now in one revolution the point moves through a distance 
2'nr so that if a point rotates uniformly at N revolutions per 
second, the velocity at a radius r is given by 

v= 27TrN (3). 

Numerical Example. // a shaft 4 inches in diameter rotates 
at a uniform rate of 80 revolutions per minute, what is the peripheral 
velocity of the shaft in feet per second ? 

In this case ? = 2 ins., N == ^ per second. 

.*. Peripheral velocity v in inches per second 

= 27rrN 

= 2 X 3-1416 X 2 X fj 
= 16-76. 

/. Peripheral velocity = ^^ 

= 1-40 feet per second. 

Centripetal and centrifugal force. If a body moves with uni- 
form velocity v feet per second in a circle of radius r feet (Fig 150), 
the velocity hodograph will be a circle of radius v, the radius XO of 





Curbed Path 



Velocity 
Hodogiaph 



Fig. 150. Centripetal Acceleration. 




Acceleration 
Hodograph 



the hodograph being at right angles to the corresponding radius 
OPq of the curved path. When the point in the curved path 
has reached P^, the radius has turned through a right angle and, 
in reaching the corresponding point 1 on the velocity hodograph, 
the radius has turned through the same angle. The velocity 
hodograph therefore turns through a complete circle in the same 
time as the body moving in the curved path completes its circle. 
The acceleration hodograph will also be a circle because it is 
obtained from the velocity hodograph by the same construction 



XV] MOTION IN A CURVED PATH 245 

as that employed for drawing the former. The radius 70 in the 
acceleration hodograph is parallel to the tangent at O and is thus 
at right angles to XO and opposite to the radius OPq of the 
curved path ; a revolution of the acceleration hodograph will also 
be completed when one revolution in the curved path is completed. 

We get therefore the result that with uniform motion in a 
circle there is a constant acceleration towards the centre. This 
acceleration is usually called the centripetal acceleration; its 
magnitude can be found as follows. 

Let t seconds be the time taken to complete the circle, then 
we have 

"=— <*)= 

also the acceleration a is the velocity on the velocity hodograph, 

^'^ (5). 



Dividing 


we get 


.*. a = 

V 

a = 


t 
r 

V 

t 
r 




• 








If the weight of the 


body is 


w, 


we have by 


the rule 




T?rkrr>*» — 


_ Weight 


X acceleration 





or o = - (6). 



9 

a constant "centripetal force" acting towards the centre of the 
circle to maintain the motion. 

The force equal and opposite to this, which is the apparent 
force acting outwards upon the body, is called the "centrifugal 
force," the two terms being often confused. 

The centrifugal force is really the force acting outwards at the 
same radius as the rotating body which will equilibrate or balance 
the system of forces acting on the body, as the examination of 
the equilibrium of such bodies is correctly dealt with by con- 
sidering the forces acting on the body together with the reversed 
radial accelerating force as forming a system in equilibrium. 

Since the weight of a body acts at the centre of gravity and 
the centrifugal force acting on each portion of the body is pro- 
portional to the weight of that body, it follows that the resultant 
centrifugal force also acts through the centre of gravity. 



246 MOTION IN A CURVED PATH [ch. 

We get therefore, from equation (6), 

Centripetal or centrifugal force 
Wa 



^ g 



Wv^ 



(7). 



gr 

If the velocity is given in terms of revolutions per minute (N) 

2'nrN 



we have, since v = 

F.= 



60 ' 

3600(7r 

\p) 



3600gr 

= -00034^^2^17 (9). 

Numerical Examples. (1) What force acting horizontally 
tends to overturn a train weighing 100 tons when running round a 
curve of 500 feet radius at 60 miles 'per hour ? 

In this case W = 100 tons, 

V = 60 miles an hour 
= 88 ft. per second, 

r = 500 ft. 
.*. Centrifugal force which tends to overturn the train 

" '~ gr 

^ 100 X 88 X 88 

32-2 X 500 
= 48 tons nearly. 

(2) At how many revolutions per minute must a stone iveighing 
I lb. whirl horizontally at the end of a string 5 feet long to cause 
a tension of 2 Ihs. in the string ? 

In this case pf = J, 

F =2 
r = 6. 
:. using equation (9) 

2 = •00034i\^2 5 j^ 

N^ = ? 

5 X -00034' 

N = 68-6 revolutions per minute. 



XV] 



MOTION IN A CURVED PATH 



247 



Applications of centrifugal and centripetal force. There are 
a large number of problems in engineering practice in which 
centrifugal and centripetal force are of importance. 

Railway Curves and Motor Tracks. 

When a railway train or motor car goes roimd a curve the 
radial acceleration induces forces which tend to overturn it, and 
this has been the assigned cause of accidents even in recent years 
— ^for instance the railway accident at Salisbury a few years ago. 
Those students who have played with model steam-engines will 
have found that when the speed gets high the engine wiU often 
fall over at a bend. 




Fig. 151. Railway Curves and Motor Tracks. 

To minimise these dangers it is now the practice to tilt or 
"super-elevate" the rails and to bank the motor track at a bend, 
the arrangement in the latter case being that the surface is 
perpendicular to the resultant of the weight of the body and the 
centrifugal force. It is commonly stated that in railway tracks 
also the surface should be perpendicular to this resultant, but 
such is not the case. In the railway track the problem is to 
give an elevation which will prevent the inner wheel from 
lifting off the rail; this means that the resultant of the weight 
and centrifugal force must act inside the tread of the outer 
rail. Referring to Fig. 151, the resultant force R is obtained by 
considering the triangle of forces ahc ; if the track is so banked 
up that this resultant acts at right angles to it, there will be no 
tendency for the body to overturn. 

Our problem therefore becomes that of determining the angle 
Q so that ^C is perpendicular to i?. 



248 MOTION IN A CURVED PATH [cjh. 

Now each side of the A ABC is perpendicular to a side of the 
A ahc ; therefore these triangles are similar and 

BC ^hc 
AB ah' 

but bc = Fc = , 

^ gr 

where v is the velocity of the train or car and r is the radius of 

the bend; also 

ah^W, 
BG _ Wf _ ?^ 
•'• AB~Wgr~ gr' 

i.e. tan^=- (10). 

gr ' 

This result gives us the angle of tilt that should be provided 
to bring the resultant force at right angles to the surface and can 
be correctly applied to the case of the motor track. 

This treatment would however, as the following numerical 
example shows, give a much higher super-elevation for railways 
than is ever adopted. 

Numerical Example. // the gauge of a railway is 4' 8J", find 
the super -elevation required for a curve of 400 feet radius at a speed 
of 60 miles an hour if the resultant force is to he perpendicular to 
the rails. 

In this case v = 88 feet per second, 

G = 56-5 inches, 
r = 400 feet, 

g = 32*2 feet per second per second, 
^ _ 88 X 88 
^"^ ^ ~ 32-2 X 400 
= •6012; 
.-. <9 = 310° approx.; 
.*. e = G sin B 

= 56-5x sin 310° 
= 29-1 inches nearly. 

Centrifugal governors or conical pendulums. The centrifugal 
governor is a device for regulating the speed of engines and motors 
and in its simplest form was employed by James Watt. A simple 
form, shown in Fig. 152, has two balls carried by arms pivoted 
to a collar A upon a shaft driven from the main shaft of the 



XV] 



MOTION m A CURVED PATH 



249 



machine. The arms cany links pivoted to a sleeve B which is 
movable up and down the shaft O, the motion being transmitted 
by a bell-crank lever (7 to a rod D connected to the throttle 
valve of the steam-engine. 

Should the speed of the engine increase, the radial force will 
increase and the balls will fly outwards and the sleeve B will rise 
and thus cut off the supply of steam until the engine has regained 
its normal speed. 

We can find the relation be- 
tween the height h and the radius 
r of the baUs for any given speed. 

The forces acting upon each 
ball are its weight W — called the 
"Controlling Force" — and the 
centrifugal force F^,* 

Since the arms are freely 
pivoted to the collar A, the 
arms will move until there is 
no tendency to move about the 
pivot. But we have seen that 
the tendency of a number of 
forces to rotate a body about 

any point is measured by the sum of the moments of the forces 
about the point, so that in the present case this moment must 
be zero, and neglecting the weight of the arm we therefore have 

. i.e. F^h = Wr, 

r "F' 




Kg. 152. 



Watt's Centrifugal 
Governor. 



Fch- Wr = 0, 



or 



_ Wgr 



h = 






[by equation 7], 



(11). 



If we wish to use a formula in terms of the number of revolutions 
N per minute we use equation (8), 

h _ 3600gr 

r 



I.e. 



].e. 



^2 _ 3600srr 

4:7Thh ' 



N 



~27r\/ 



(12). 



250 MOTION IN A CURVED PATH [ch. 

Numerical Example. Find the speed at which a simple 
centrifugal governor will run when the height is 9 inches and find 
the amount by which the balls will rise when the number of revolu- 
tions per minute increases by 5. 

In this case ^ = 9 inches = -75 foot ; 



^60 /32;! 

27tV -75 
= 63 revolutions per minute nearly. 



If iV^ = 66 we shall have 

^^ 60 /32^ 

^.^gg,^ 60^x322 



h = 



4:7T^h ' 

60 X 60 X 32-2 



66 X 66 X 4772 
= '674 foot nearly 
= 8-09 inches. 
.*. The balls rise by 9 - 809 

= '91 inch. 

Balancing rotating parts. If a wheel or other revolving body 
has its centre of gravity out of the centre of rotation, then the 
whole body may be considered as a weight concentrated at its 
centre of gravity and thus rotating in a circle whose radius is 
equal to the distance from the centre of gravity to the axis of 
rotation. 

The resulting radial force may at high speeds cause severe 
vibrations and will interfere with smooth running besides causing 
heavy stresses upon the shaft and bearings. 

Rotating bodies which give rise to these centrifugal forces 
are said to suffer from want of balance and the problem of 
removing these forces and similar forces caused by rotating parts 
is called "balancing." In many cases this problem is an ex- 
ceedingly difficult one, and in some cases of engines in electric 
power stations which caused severe vibrations in adjoining 
buildings great expense and inconvenience have resulted, due to 
the inability of even the leading authorities to quite remove the 
lack of balance. 

We cannot at the present stage go fully into the more advanced 



XV] MOTION IN A CURVED PATH 251 

aspects of the problem but the numerical example given below 
will indicate that a small divergence of the centre of gravity 
from the centre of rotation may cause quite serious forces. 

Numerical Example. A flywheel weighing 5 tons has its centre 
of gravity -^ of an inch from the centre of the shaft. Find the force 
upon the shaft caused by the lack of balance when running at 200 
revolutions per minute. 

In this case we may use equation (9), thus getting 

F, = -00034/^17 

= -00034 X 200 X 200 X tIt7 X 5 
= -57 ton. 

If we wish to balance an unbalanced body we may add a 
weight to it at such a point that it will cause a centrifugal force 




Fig. 153. 

equal and opposite to that caused by the eccentricity. Suppose 
for instance that a body of weight W is rotating about an axis 
0, Fig. 153, and that its centre of gravity (r is at a distance 
e from 0. 

This causes a centrifugal force F^. which may be balanced by 
an equal and opposite force Fc caused by a weight w placed at 
radius r at a point diametrically opposite to G. 

Now this means that we shall bring the centre of gravity of 
the whole body including the weight w back to O. 

Therefore by moments about O we shall have 

wr = We, 
,% w= — ^. . .(13) 



252 MOTION IN A CURVED PATH [ch. 

We can get the same result by considering the centrifugal 
force Ff. . 

We then have F, = '00034:N^eW 

i.e. no = eW, 

We 
or w = — as before. 

r 

Projectiles. If a body such as a stone or a bullet is projected 
into the air in a direction other than vertical it describes a curved 
path called the trajectory which, as we shall show later, is a 
parabola if the resistance of the air is not taken into account. 



2 «Jk 3 

Fig. 154. Projectiles. 

Suppose that a body is projected with a velocity u from a 
point P, Fig. 154, at an inclination 6 with the horizontal. Then 
this velocity may, as we have seen before, be resolved into 
a vertical component u^ and a horizontal component % whose 
values can easily be found by drawing the A 123 to scale or can 
be found by trigonometrical calculation as follows : 

u^, = u sin 6 (1), 

UJ^ = u cos 9 (2). 

Now the on! 7 force acting upon the projectile, if air resistance 
is neglected, is that of gravity which acts vertically downwards 
and will give a vertical downward acceleration to the body. 
But this vertical force can have no effect upon the horizontal 
component of the velocity, so that the horizontal component Uj^ of 
the velocity remains constant. 

Now suppose that after a time t the projectile has reached 
a position Q, the components of its velocity v then being u„ and UJ^. 



XV] MOTION IN A CURVED PATH 253 

Then the vertical distance y will be the same as would be 
obtained by projecting a body vertically with velocity u^ so that 
by formula (6) on p. 96 we have 

y = ujt- \gt^ (3), 

ic = uj,t (4). 

Putting t = — ,UJ^ being a constant, we have 

but this is of the general mathematical form 

y = ax + bx^, 

where a and b are constants, and we know that the corresponding 
curve is a parabola. 

Therefore we have proved that the path of the projectile 
is a parabola. 

We now wish to obtain some way of finding the height h to 
which the projectile will rise and the range dj i.e. the distance 
away from P at which the body will again be at the same 
level. 

The height h will be the height to which a body will rise when 
projected vertically upwards with a velocity u^. 

From equation (12), p. 97, we get 

* = %' (6). 

The time to reach this point is given by 

1 = "^ (7). 

g 

By the time the body has arrived at T it has been moving 
for a time 2t and since it has been moving in a horizontal direction 
with a constant velocity % , the range d must be given by 

d = constant velocity x time 
«?^^iLJ^ ^gj 



254 



MOTION IN A CURVED PATH 



[oh. 



When we are calculating with the aid of trigonometry we can 
make results of equations (I) and (2) as follows; 

2u sin 6 . u cos B 



d^ 



W2 



. 2 sin ^ cos B 



.sin 2^ (9). 



Direction to give greatest range for a given velocity. If the 
velocity is fixed the range for different directions varies as sin 26 
as given by equation (9). Now the greatest value of the sine of 
an angle is 1 and occurs when that angle is 90°. 

Therefore the maximum range occurs when 29 = 90° or 
6 = 45°, so that to send a projectile the farthest distance 
horizontally we should project it at an angle of 45° to the horizon. 

Projectiles considered from the hodograph. With a pro- 
jectile, the acceleration is, as we have seen, constant and is 
vertically downwards and the horizontal velocity is constant so 
that in equal times the body moves through equal horizontal 
distances. 

Velocity Hodograph 






t? 


P2 

1 


Trajectory 

i3 R • R 


PoZ 


— ^- — -^ 






~"^ 


1 



Fig. 165. Hodograph for Projectiles. 

We saw on p. 243 that the acceleration is the velocity with 
which the point moves on the velocity hodograph so that as 
this is constant the points on the velocity hodograph are at equal 
distances apart. This gives us the velocity hodograph shown 
in Fig. 155. Working backwards from this and the knowledge 
that in equal times the horizontal distances are equal, we draw 
a number of vertical lines at equal distances apart and draw 
PqPj parallel to XO, then PiP^ parallel to Zl and so on. This 
wUl be recognised as the link and vector polygon construction 
which gives a parabola when the points are near enough together. 



xv] 



MOTION IN A CURVED PATH 



255 



Consider for instance the graphical construction for the b.m. 
diagram of a beam carrying a uniform load. 

Numerical Examples. (1) A shot is projected horizontally 
from the top of a tower 60 feet high with a velocity of 200 feet per 
second. After what time will it strike the ground and how far away 
from the base of the tower will it then be ? 

In this case the trajectory will be somewhat of the form shown 
in Fig. 156 and the time taken will be the same as that taken to 
fall 50 feet from rest. 



•^^^ 




'^P77^77r 7Z77WW///W/V?'>//-/WWW'///W/'/^//''/''''^/7^ ^ 



Fig. 156. 



Therefore we have from equation (9), p. 97, 

•• ^ ~ g ""32-2"'^^^' 

t = VzTl = 1-76 seconds, 
.-. eZ -= 200 X 1-76 = 352 feet. 

(2) // a man can throw a stone 90 yards, how long is it in 
the air, and to what height will it rise? 

In this case <Z = 90 yards 

= 270 feet. 

And we have seen that 6 = 45° for maximum range so that 
u„ = %. 

2Uy . UJ^ 2uJ^ 



.*. smce 



d = 



9 



-J 



gd 

2 



270 X 32-2 



65-8 feet per second ; 



266 MOTION IN A CURVED PATH [oh. 

.-. from (6) ^ = ^ = I = 67-5 feet, 

from (7) Time to top = — , 

.„,,.. 2u^ 65-8 

/. Total time = — ^ = ;r-^ 

g 161 

= 4*1 seconds nearly. 

SUMMARY OF CHAPTER XV. 
The motion of bodies moving in a curved path is conveniently 
studied by a graphical construction called the Hodograph. 

The angular velocity («) of a body rotating about a fixed axis is 
equal to the angle through which the body rotates round the axis in 
a unit of time. 

... f) = o)r = 2nrN. 
Centripetal acceleration =— . 

Centripetal or centrifugal force = 



Governors : 



^•00034:N2rW, 



v^ 

Projectiles. The path of a projectile is called its trajectory and 
if air resistance is neglected it will be a pai*abola. The horizontal 
component of the velocity of a projectile is constant. 

The vertical component has gravity acceleration acting against it. 

Range of projectile ==d= — sin 20. 

The greatest range on a horizontal plane for a given initial velocity 
occurs when the angle of projection is 45°. 

EXERCISES. XV. 

1. A body weighing 2 tons moves in a circle of radius 10 ft. 
6 ins. maldng 180 revolutions per minute. Find its kinetic energy 
in ft. -lbs. 

2. A weight of 1 lb. is fastened to the end of a string 3 ft. long 
and made to perform 50 revolutions per min. with uniform velocity, 
the revolutions taking place in a horizontal plane. 

Determine the tension of the string. 



XV] MOTION IN A CURVED PATH 257 

3. Find the speed at which a simple Watt governor runs when 
the arm makes an angle of 38° with its vertical. Length of arm 
from centre of pin to centre of ball =18 inches. 

4. A railway carriage of weight 2 tons is moving at the rate 
of 60 miles per hour on a curve of 770 ft. radius. If the outer rail 
is not raised above the inner, find the lateral pressure on the rail. 

5. A string 4 ft. long which can just support a weight of 9 lbs. 
without breaking is placed on a horizontal table. To one end is 
fixed a weight of 8 lbs. and the free end is held and the weight is 
swung round. Find how fast the weight may go so as just not to 
break the string. 

6. At what speed must a locomotive be running on level lines 
with a curve of 968 ft. radius if the thrust on the rails is ^ of its 
weight ? 

7. A locomotive engine weighs 38 tons and travels round a 
curve of 800 ft. radius at 50 miles per hour. Find the centrifugal 
force. Show how to find the direction and magnitude of the 
resultant thrust on the rails due to its weight and the centrifugal 
force. 

8. A motor car moves at constant speed in a horizontal circle 
300 ft. radius. The track is at 10° to the horizontal. The plumb- 
line makes 12° with what would be perpendicular if the car were 
on the horizontal. Find the speed of the car. 

9. A flywheel 5 ft. 3 ins. in diameter has a rim weighing 
1000 lbs. Find the number of foot-pounds of work required to set 
this rotating 120 times per minute. 

10. A brake wheel 4 ft. in diam. on a horizontal axle is furnished 
with internal flanges which, along with the rim, form a trough 
containing cooling water. What is the least speed which will prevent 
the water from falling out? 

11. Find the greatest range which a projectile with an initial 
velocity of 1600 ft. per sec. can attain on a horizontal plane. 

12. A rifle has a range of 1000 yards. What would the range 
be under the same circumstances if fired in the moon where the 
force of gravity is ^ that of the earth ? 



A. M. 17 



CHAPTER XVI 

MECHANISMS 

For our present purpose we will regard a "mechanism" as 
a device for transferring motion from one point to another in 
a machine. In many cases the kind of motion becomes changed 
in the transformation, for instance a rotation becomes changed 
into an oscillation or a reciprocation or vice versa. The name 
" linkage mechanism " is used for those mechanisms in which rods 
are employed which are pivoted together, such rods being called 
links or elements, and the whole collection of rods being called 
a " kinematic chain." 

Crank and Connecting-rod or Steam-Engine Mechanism. 
This is about the most common linkage mechanism employed in 
machinery, and it is used for converting a reciprocating motion 



Connecting Rod 




2 p 

Fig. 157. Crank and Connecting-rod Mechanism. 

into a rotary motion or vice versa. It is used on nearly all 
steam, oil or petrol engines, in which the reciprocation of the 
piston is converted into a rotation of the shaft, and in a very 
large number of mechanical presses in which it is employed to 
convert the rotary motion of a shaft into the reciprocating motion 
of a press-head. 



CH. XVl] 



MECHANISMS 



259 



The mechanism consists of a link AB, Fig. 157, called the 
cranky which is fixed to a rotating shaft and is pivoted at its end 
^ to a rod BG called the connecting-rod. The connecting-rod is 
pivoted at its other end O to a block E called the cross-head 
which is guided so as to move in a straight line and is connected 
by a piston-rod to the piston D of the engine. On the rotation 
of the crank the cross-head is caused by its guides to reciprocate. 
It is interesting to note that James Watt did not use this 
mechanism for his steam-engine because one of his workmen had 
stolen the idea and obtained prior patent rights for it. He 
devised what is called "the Sun and Planet mechanism" which 
is practically never used nowadays, and the consideration of 
which is outside our scope. 

In this and all other mechanisms to be described the student 
must trace out the movement by actually drawing the mechanism 
to scale in a number of its possible positions, or else by making 
a model of the mechanism and attaching a pencil to the point 
whose motion he wishes to study. The pencil will then trace 
out on a piece of paper the path in which that particular point 
moves. Such models can be very easily made by the aid of the 
constructional toys now on the market. 

Velocities in Mechanism. Instantaneous or Virtual Centre. 
Suppose that a body as shown shaded in Fig. 158 is moving 




Fig. 168. Instantaneous or Virtual Centre. 

in any manner and the velocities v^, Vg of two points A and B 
in it are known in magnitude and direction. Draw AI per- 
pendicular to v^, and BI perpendicular to v^, then the intersection 

17—2 



260 MECHANISMS [ch. 

/ is called the instantaneous or virtual centre because A and B 
may both be regarded as rotating for the instant about this 
point. We may therefore study the motion of the body at the 
particular instant under consideration by imagining it to be 
rotating about the point /. It is important to remember that 
unless the body is rotating about a fixed point, / will be constantly 
changing and the curve in which / moves is called the "centrode." 
At any instant, however, we can find the relation between the 
velocities of the various points of the body if we know the 
instantaneous centre, because when a body is rotating we have 
seen that the velocity of any point in it is proportional to the 
radius of the point. We therefore have 

vb Br 

To obtain the velocity of any other point, say C, we join CI and 
draw a line at right angles to it. This gives the direction of 
Vq, and its magnitude is given by the relation 

Application to Crank and Connecting-rod Mechanism. Sup- 
pose that the shaft A, Fig. 159, is rotating uniformly so that the 
crank pin B has a uniform velocity v^, at right angles to the 
crank. Draw CI perpendicular to the direction of the cross-head 
and produce AB to meet it at /. Then / wiU be the instantaneous 
centre of the movement of the connecting-rod BC. Because BI 
is at right angles to v^, we therefore have 

'i-% »>. 

but since the triangles BIC and BAD are similar we have 

CI ^AD 
BI AB' 
Vq AD 
" Vb AB' 

that is to say Vq = -j?^ . AD (2). 

If therefore we choose our scale of velocity so that AB 
represents Vr the length of the crank pin AD will give us the 
velocity of the cross-head and therefore of the piston to the same 
scale. By repeating this construction for a large number of 



XVI] 



MECHANISMS 



261 



positions of the crank pin and cross-head we can find the velocities 
in the different positions, and from these we can draw a diagram 
showing the manner in which the velocity varies. Two con- 
venient forms of diagrams are shown in the figure. One is drawn 
upon a base of the stroke and is obtained by projecting the point 
D upon the line CI, thus obtaining the point F, and joining up 
points such as F. This diagram is useful when we wish to find 
the velocity for a given position of the cross-head or piston. The 




Fig. 159. Velocity Diagrams. 

other form shown is called a polar diagram and is extremely useful 
for finding the velocity of the piston for different positions of the 
crank. It is obtained by drawing with centre A an arc of radius AT) 
to meet the crank (produced if necessary) at E and joining up the 
points thus obtained. It wiU be found to give two loops as shown. 
In the use of these diagrams for any position of the cross -head, 
say C\ the velocity of the cross-head is given by C'F' or by AE'. 
Force in connecting-rod; crank effort. The force Q in the 
connecting rod can be found by drawing the triangle of forces 
1, 2, 3 as indicated in Fig. 157. This force Q can be resolved 
into a component T along a component J at right angles to the 
crank . The force J is called the ' ' crank effort. ' ' If no work is lost, 
the work done by P per second must be equal to that done by J , 

/. J.Ve=P.Vc, 



or 



J = 



Vb 
P AT) 
= ^ (Fig. 159) 



(3). 



262 MECHANISMS [ch. 

Watt's Parallel Motion. This mechanism was used by James 
Watt to guide the valve rods of his beam engines without the 
necessity of providing a cross -head and was regarded by him as 
one of the most ingenious of his inventions. The rod AB, 
Fig. 160, is pivoted at the point A and is connected by a " coupler " 




Fig. 160. Watt's Parallel Motion. 

BC to a rod CD pivoted at the opposite side as shown. A point 
E is taken on BC such that 

BE ^GD 
EC~ AB* 

and as one or other of the rods ^J5 or CD is oscillated the point 
E will be found to move in a line which is for all practical purposes 
straight for small amounts of oscillation. For a complete 
revolution of AB the point E will be found to trace out a looped 
figure as shown. 

Slotted lever quick- return mechanism. This mechanism is 
used to reciprocate the ram which carries the cutting tool of 
shaping machines and it has the property that the time taken in 
the forward or cutting movement of the ram is greater than the 
time taken in the return or idle movement; this is economical 



XVIJ 



MECHANISMS 



263 



because it reduces the time during which the cutting tool is doing 
no useful work. This mechanism is often called the " Whitworth 
quick-return motion/' but this description is not quite correct, 
Whitworth's mechanism being slightly different although pos- 
sessing the same property. 

The mechanism consists of a lever AB (Fig. 161) pivoted at 
the lower end to a fixed point A and provided with a slot C in 
which works a crank pin D which rotates about a centre E 
vertically above the fixed point A. The upper end B of the 




— « — >. I — •» 



iiiiiiimiiiiiiii!iuiiiii}l^)h 



f Stroke of Tool^, 



Fig. 161. Quick-return Mechanism. 



slotted lever is connected by a rod BF which is pivoted at the 
end F to the ram of the shaping machine. This ram runs in 
horizontal guides and it is usual to provide means for adjusting 
the position of the crank pin D so as to alter the stroke of the 
ram. The mechanism is shown diagrammatically in the figure in 
its extreme position. While the crank pin is moving through 
the arc DXD' the tool ram moves from F to F' thus effecting the 
cutting stroke and while the pin moves through the arc D'YD 
the ram makes the idle or return stroke. If the crank pin rotates 
with imiform velocity, the time taken on the cutting stroke must 
be proportional to the length of arc DXD' and on the return 
stroke to the arc D'YD; and since the arcs are proportional to 
the angles that they subtend at the centre of the circle and 
therefore are also proportional to the halves of such angles, the 



264 



MECHANISMS 



[CH. 



cutting and return times are respectively proportional to the 

angles DEX and DEY. 

, cutting time /_DEX 

:. we have — 7 — —-. = , rirrp- • 

return tmie /_DEY 

But since the mean cutting speeds are inversely proportional 
to the corresponding times we have 

mean return speed cutting time 



mean cutting speed 



return time 

LDEX 

/_DEY' 



It should be noted that this ratio deals only with the average 
or mean cutting or return speeds, because the actual speed varies 
at different points of the stroke. 

Numerical Example. In a Whitworth quick-return gear of a 
shaping machine the stroke is 8* and the ratio of home and cutting 
strokes is 3:5. The line of 
stroke of the ram produced passes 
through the extreme positions of 
the connecting-rod pin at the 
end of the slotted lever. If the 
distance between the centre of 
the driving plaie and the axis 
about which the slotted lever oscil- 
lates is 6*, find the crank radius 
and length of the lever. 

Fig. 162 shows the arrange- 
ment in this case; BB' is the 
stroke which is given as 8 ins. ; 
we are also given 

cutting time _ 3 
return time 5 * 

. /_DEY 
'' /_DEX 

/_DEY 




Fig. 162 



that is 



or 



Z.DEY+ Z.DEX 
/_DEY 



\m 



5+3 

3 

8' 



3 

8' 



l,DEY = f X 180° = 67-5^ 



XVI] 



MECHANISMS 



265 



Now ^DAE = 90° - /LDEY = 22-5°, 

and /_ABK = 90° - 22-6° = 67-5.° 

We are now in a position to draw the figure to scale, and 
by first drawing BB' horizontal to represent Sins, and then 
LB'BA and lBB'A each = 67-5° we get the point A and AB 
which is the required length of the lever and will be found by 
measurement to be about 10'5 ins. 

Now set up AE = 6 ins. and draw ED perpendicular to AB. 

Then ED is the crank radius and will be found to be about 
2-29 ins. 

If as is preferable we proceed by trigonometrical calculations 
we shall have 

'^'^ = sin 22-5°, 

4 



BK 
AB 

AB = 



225 



= 1046 ins. 



Also 



|f = sin22-5'. 



.-. ED = EA sin 22-5° = 6 sin 22-5° = 2-29 ins. 
Toggle Mechanism. The name "toggle" is used to denote 
a linkage mechanism in which one part receives a very small 




Fig. 163. Toggle Mechanism. 



266 



MECHANISMS 



[CH. 



motion while another receives an appreciable movement; it is 
used as a means of exerting heavy pressures in presses and is 
also used in a large number of every -day appliances, such as the 
devices which are to be found for closing bottles of various kinds. 
Fig. 163 shows a common arrangement for use in mechanical 
presses. The toggle links ABy BC are connected at one end A to 
a fixed support and at the other end C to the press-head. The 
joint B is connected to an eccentric D carried by a rotating shaft 
E the arrangement being such that as the shaft rotates the 
eccentric is reciprocated and a small movement is given to the 




F 
— ^ 




2 4 1 



Fig. 164. 



Fig. 165. Forces in Toggle 
Mechanism. 



press-head which exerts a very considerable pressure. Fig. 164 
shows diagrammatically one form of toggle closing device for 
stoppered bottles. A wire loop AB passes over a groove on top 
of the stopper X and is pivotally connected at the point B to 
a bent wire lever CD which is pivoted at the point C to a wire 
ring fastened round the neck of the bottle. As the point D is 
moved about the centre C in the direction of the arrow the loop 
AB is pulled downwards and exerts a strong closing action upon 
the stopper. 

We can examine in the following manner the pressure exerted 
in the toggle press for any position of the toggle levers. Referring 
to Fig. 165 the force or effort F exerted by the eccentric is resolved 



«n 



XVI] MECHANISMS 267 

into two forces Q acting down the toggle links, and these forces 
Q can be resolved into vertical components R one of which at 
the point B is carried by the framing of the machine and the 
other of which is the pressure exerted on the press-head at C. 
The A 123 is the triangle of forces and from this we get that 

I; 4, 1 

2 

= cot 6, 

i.e. R = -^cotd 

A glance at the trigonometrical tables will show that as 
an angle gets small its cotangent increases very rapidly so 
that we see that if the angle 6 is small the pressure R will be 
very many times more than the force F. In the limiting con- 
dition ^ = 0, the pressure would theoretically be infinitely great, 
but ia the practical use of the mechanism 6 can never be exactly 
zero although it may be very near to it, as in practice there is a 
limit to the pressure which the mechanism can exert owing to 
the yielding of the various parts composing it. 

Cams and Wipers. Cams, or wipers as they are sometimes 
called, are a form of mechanism for converting rotary motion 
into a reciprocating or oscillating motion. 

Fig. 166 shows a cam C for giving a reciprocating motion to 
a shaft E from a rotating shaft A. The shaft E is guided by 
a slide G and carries at its end an anti-friction roller D which 
rides upon the face of the cam. A spring F is employed for 

o 




F 

^ ^ l->_Z__J ** <=?H 

A 

Fig. 166. Cams and Wipers. Kg. 167. Cams and Wipers, 

keeping the roller in contact with the cam when the latter is in 
such a position that the shaft E is moving towards the shaft A. 
Other devices for this purpose are sometimes devised, a common 
one being that the roller runs in a groove cut in the cam disc 
and is thus positively moved in both directions. The form 
shown in the figures is, however, usually preferable. 

In Fig. 167 is shown a cam G communicating an oscillating 



268 



MECHANISMS 



[CH. 



motion to a lever E pivoted at F. A rod G pivoted at the other 
end of the lever communicates the motion to the required part 
of the machine, and a spring H keeps 
the roller D in contact with the cam. 
Cams of this kind are used in almost 
every form of gas-engine for operating 
the valves in the required sequence. 
The cams shown in Figs. 166, 167 
are often called plane or edge cams, 
the form shown in Fig. 168 being 
called a surface or drum cam. In the 
latter case the cam is formed as a 
groove in a drum C carried by a 
rotating shaft A. A roller carried 
by a lever D engages the groove and 
the lever is pivoted at E and con- 
nected at its extreme end to a slide 
F to which the cam communicates 
the required motion. The lever is 
stationary while the centre of the 
groove remains in a plane section 
of the drum normal to the axis. In 
the case of plate cams the slide or 
lever is stationary for the portion of 
the cam that is concentric with the 
axis of the shaft. 




f///////////// 



Fig. 168. 



Surface or Drum 
Cam. 



Design of a Plate Cam. A plate 
cam for giving a reciprocating move- 
ment in a straight line passing through the centre of the cam 
shaft can be designed as follows. Suppose that we are given the 
following particulars : 

The diameter D of the cam shaft. 

The diameter d of the roller. 

The minimum thickness t of the cam. 

The lift or height h of the movement of the roller. 

The manner in which the roller has to rise and fall ; this is 
called the "timing" of the cam. 

To make our illustration more clear we will assume that our 
cam is required to move the slide uniformly upward during 



XVI] 



MECHANISMS 



269 



one-third of a revolution, is then required to remain stationary 
for another third of a revolution and has finally to fall uni- 
formly during the remainder of the revolution. 

First draw with centre Oa circle of diameter D (Fig. 169) to 
represent the cam shaft, and draw the line OX along which the 
roller has to reciprocate. Next make CD equal to ty the minimum 
thickness of the cam and find the centre A of the roller in its lowest 
position. AB is next set up equal to the lift h and AB is divided 




Fig. 169. 

up into a convenient number of equal parts, say 4. A number of 
equally spaced radial lines 01, 02, etc. are then drawn, 12 being 
taken as a convenient and sufiicient number. 

In moving through one-third of a turn, i.e. from OA to 04, 
we have to rise a height h and have to do so uniformly; we 
therefore make 01 = 01 and 02 = 02 ; 03 = 03 and 04 = 04 as 
indicated by the circular arcs shown in dotted lines. During the 
next one-third of a revolution the roUer has to remain stationary 
so that we draw an arc with centre from 4 to 4', and since the 



270 



MECHANISMS 



[oh. 



roller has to fall uniformly during the remainder of the revolution 
we repeat the construction for the points 1, 2, 3, 4 to get the 
points 1', 2', 3', 4'. 

By joining up the points thus obtained we obtain the curve 
shown in chain dotted lines, this curve being that for the centre 
of the roller. We then go round the curve and draw the roller 
circle all round it to give the effect of the roller running along 
such a curved path. By drawing a line to touch the roller on 
the inner side in all its positions we get a curve which a mathe- 
matician would call an envelope and which gives us the shape of 
the cam required. 

Pawl and Ratchet Mechanism. This form of mechanism is 
employed for giving an intermittent motion from a continuous 
motion and is most commonly employed for converting an 
oscillating motion into an inter- 
mittent rotary motion. The 
ratchet wheel A, Fig. 169 a, has 
teeth on it which are adapted to 
be engaged by a pawl or "click" 
B carried by a pivot D on an 
oscillating lever G. The pawl 
drives when the lever is moved 
in the direction of the arrow and 
slips over the ratchet when moved 
in the opposite direction and re- 
engages one of the teeth beyond, 
when it reverses again. The movement given to the pawl should 
be a whole number of times the distance apart of the teeth. To 
prevent the ratchet wheel from moving back with the pawl, a 
stop-pawl Ey pivoted upon a pin F, is often provided. 

The ratchet brace gives a very familiar example of this 
mechanism. 




Fig. 169 a. Pawl and Ratchet 
Mechanism. 



XVI] MECHANISMS 271 



SUMMARY OF CHAPTER XVI. 

The crank and connecting-rod mechanism is a device for convert- 
ing a reciprocating motion into a rotary motion and vice versa. 

The virtual centre of a body moving in any manner is a point 
about which the body may be regarded as rotating at any particular 
instant. 

The path moved through by the virtual centre is called the 
centrode. 

Watfa parallel motion. The point which has to move in a straight 
line divides the coupler inversely as the lengths of the pivoted rods. 

Slotted lever mechanism. The return or idle motion is quicker 
than the forward or cutting motion, the relative values are obtained 
by considering the angles turned through by the crank pin in the 
extreme positions. 

Toggle mechanism is employed to obtain a heavy pressure 
moving through a small distance from a small force moving through 
a greater distance. The forces are obtained by the ordinary triangle 
of forces. 

Cams are a device for giving an oscillating or reciprocating motion 
from a constantly rotating shaft. They may be " plane or edge " 
cams or '* surface or drum " cams. 

Ratchets are a form of mechanism for obtaining an intermittent 
motion from a constant reciprocating or rotary motion. 



» EXERCISES. XVI. 

1. The connecting-rod of an engine is 2^ times the stroke in 
length. Find graphically (a) the position of the crank when the 
piston is at half -stroke, {h) the position of the piston when the crank 
is 90° from a dead point. 

2. The stroke of an engine is 2 ft. and connecting-rod 4 ft. 
long. The thrust on the piston is 12,000 lbs. When the crank 
is vertical find 

(1) The thrust on the cross-head. 

(2) The thrust on the connecting-rod. 



272 MECHANISMS [ch. xvi 

3. The connecting-rod of a steam-engine is 4-5 ft. long and 
the crank has a radius of 1-5 feet. Draw a curve showing the 
displacement of the piston for different angular positions of the 
crank. 

4. In a steam-engine mechanism the crank radius is 10 inches 
and the connecting-rod 10 inches long. If the crank makes 120 
revs, per min. find the velocity of the piston when the connecting- 
rod is at right angles to the crank. 

5. Trace the curve drawn by the Watt parallel motion if the 
length of the pivoted links is 24* and the coupling-rod 12", the upper 
pivoted link having an oscillation 20° from the horizontal. 

6. In a quick-return motion for a shaping machine the length 
of the lever is 10*46 ins. and the crank radius 1-15 ins., the distance 
between the centre of the rotating shaft and the pivot of the slotted 
lever being 6 ins. Find the ratio of return to cutting speeds and 
the stroke of the cutting tool. 



CHAPTER XVII 

BELT, CHAIN, AND TOOTHED GEARING 

Belt, chain and toothed gearing is a form of mechanism for 
converting a rotary motion about a certain centre into a rotary 
motion about another centre. In the case of a belt, the power 
is transmitted through the friction between the belt and the 
pulleys, and in the case of toothed gearing the power is trans- 
mitted through the stresses in the material of the teeth. Chain 
gearing is similar to belt gearing except that in place of the 
friction drive we have positive drive between the teeth of the 
sprocket wheels and the links of the chain. 

Belt Gearing. If the belt transmits motion from a shaft X 
to a shaft Y the pulley on the shaft X is called the driver and 
that on the shaft Y is called the follower. The belts may be 
ojpen as (a) (Fig. 170) or crossed as (6). In the open arrangement 
the driver and follower rotate in the same direction, whereas in 
the crossed arrangement they rotate in opposite directions. 
The power that can be transmitted by the gear depends upon 
the friction between the belt and the pulley, and this friction 
depends on the angle subtended at the centre of the pulley by 
the arc of contact. 

In the crossed arrangement this angle is greater than in the 
open arrangement so that in this respect the crossed arrange- 
ment is better than the other. The tension T^^ in the belt on 
the side as it comes on to the driver is greater than the tension 
Tg on the other side. If the driver is Dx feet in diameter and 
makes Nx revolutions per minute and the follower is Dy feet in 
diameter and makes Ny revolutions per minute, the work done 
against the tension T^ on the tight side per minute will be equal 

A.M. 18 



274 BELT, CHAIN, AND TOOTHED GEARING [ch. 

to the force multiplied by the distance moved by the belt per 
minute 





Fig. 170. Belt Gearing. 
Similarly the work done on the pulley by the belt on the slack 



side 



= T..7rDj,N^, 



.', total work done per minute on belt 



H.P. 



'J.2. J work done per minute in ft.-lbs. 
transmitted « 33-^^^^^ 



33.000 ^^^• 

In the absence of other information T^ may be taken as 

twice 7^2- 

When the diameter of the belt is appreciable compared with 

the diameter of the pulley for calculations Dx should be measured 

to the centre of the belt. 



XVII] BELT, CHAIN, AND TOOTHED GEARING 275 

Numerical Example. Find the h.p. that can he transmitted 
by a pulley Sft. 6 ins. diameter running at 120 revolutions per 
minute by a thin belt 6 ins. wide if the permissible tension in the 
belt is 80 lbs. per in. of loidth and the tension on the slack side is 
equal to half that on the tight side. 

In this case T^ = 6 x 80 = 480 lbs., 

^2 = 240 lbs. 

/. Ti - ^2 = 240 lbs. 

., . , 7r.3-5x 120x240 ^^ 
.*. H.P. transmitted = = 9*6. 

Velocity Ratio in Belt Gearing. Now by exactly similar 
reasoning applied to the follower instead of the driver we should 
get 

H.p. transmitted = "^gg^^^^ ^- (2), 

and if no power is lost these two must be equal. 

.-. D^Nj, = DyNy, 
no. of revolutions per min. of follower 



.*. velocity ratio = 



no. of revolutions per min. of driver 



N^ Dy' 

_ diameter of driver 
^ diameter of follower ^ '' 

We could have obtained this result rather more simply without 
going into the question of h.p. Unless the belt slips on the 
pulley, the length of the belt passing on to the driver per minute 

= circumference of driver x no. of revolutions per minute 

But unless the belt stretches this must be exactly equal to 
the length of the belt passing on to the follower per minute 

= TrDyNy, 

:. TT.DxNx^TT.DyNy, 
:. DxNx = DyNy. 

Numerical Example. A shaft running at 120 revolutions per 
minute carries a belt pulley of 3 ft. 6 ins. diameter. What must be 
the diameter of the pulley on the shaft driven by the belt if it runs 
at 300 revolutions per minute ^ 

18—2 



276 BELT, CHAIN, AND TOOTHED GEARING [ch. 



In this 


case 


Nj,= 


120, 


Dx= 3-5andiVir = 


300, 






• 
• • 


since Nx^x 


= NyDy, 












SOOjDi. 


= 120 X 3-5, 












:.Dy 


120 X 3-5 
300 
= 1-4 feet. 





Belt Speed- cones. In belt-driven machines it is often desirable 
to vary the velocity ratio transmitted, i.e. to vary the speed of 
the machine. This is usually effected by speed-cones which 
consist of two sets of pulleys whose sizes are so arranged that the 
belt will run tightly between any opposite pair. 




Fig. 171. Belt Drive for Lathe Headstock. 

Fig. 171 shows an arrangement commonly employed for 
driving a lathe. A cone of three pulleys B is mounted in the 
headstock K of the lathe and an overhead shaft L carries a 
corresponding cone of pulleys A. When the belt is between the 
pulleys 1, the driver is larger than the follower and we then have 
the quickest speed of the headstock spindle; when the belt is 



xvn] BELT, CHAIN, AND TOOTHED GEARING 277 

in the position 2 shown, the two pulleys are about equal in 
diameter so that the speed is less; whereas when the belt is 
between the pulleys 3 the driver is smaller than the follower so 
that the headstock spindle is driven at its lowest speed. 

Belt-striking gear. Fig. 171 shows also one form of device 
used for starting and stopping a machine driven by belt gearing. 
The overhead shaft is driven by a belt N and two pulleys C, D of 
equal diameter are placed alongside on the shaft. The pulley 
C is keyed to the shaft, and is called the "fast pulley," and the 
pulley D is loosely mounted and is called the "idle or loose 
puUey." The belt N passes between forks E carried by a sliding 
rod F which is moved lengthwise by a slotted lever which is 
moved to one or other of its extreme positions by means of 
chains G. 

In the position shown, the belt N is upon the fast pulley C so 
that the headstock is being driven. If the left handle G be 
pulled down the rod F will be moved to the left, and the belt 
N will be moved on to the idle pulley D which does not drive 
the shaft L because it is not keyed to it. 

Sizes of Cones for keeping Belt taut. As we have already 
indicated, it is necessary that the diameter of the pulleys in a 
cone shall be such that the same length of belt will run taut over 
all of them. 

Open belts, li Sis the distance apart of the shafts, the length 
of an open belt is given approximately by the formula 

J = 2S+|(Z)^+i)p) + <^^^^' (4). 

If therefore we are given S, and the diameters Dj and Dy 
of one pair of pulleys, the diameters of the others should be 
chosen so as to keep I practically constant. 

Crossed belts. In this case it can be proved that the length 
of belt is constant for a fixed value of aS^ if the sura of the diameters 
of the pulleys is constant so that it is quite an easy matter to 
choose suitable diameters of a cone of pulleys to work with 
crossed belts. 

Belt Reversing Gear. In some machines, such as planing 
machines, it is necessary to reverse periodically the direction of 
rotation of the working parts. With a belt drive this ca-i be 



278 BELT, CHAIN, AND TOOTHED GEARING [ch. 

effected by the arrangement shown in plan in Fig. 172 which 
we will describe with reference to a planing machine. 

The main driving shaft B carries a broad pulley A upon which 
are carried a crossed and an open belt. The driven shaft D 
carries two outside idle pulleys and a central fast pulley. Belt forks 
C are provided and are so spaced that one belt is on the fast 
pulley and one on an idle pulley. In the position shown the 
open belt is driving and the cross belt is idly rotating its pulley 
in the reverse direction. When the planing machine later reaches 
the end of its stroke, tappets or blocks adjustably mounted upon 
it strike arms which communicate their motion to the shaft 
carrying the belt forks C. The latter are then moved (upward 
on the drawing) so that the crossed belt comes on the fast pulley 



Lo I 



rii 



Ij idle 




"I ii crossed [j C | I 



6 D 

Fig. 172. Belt Drive for a Planing Machine. 

and the open belt moves on to the upper idle pulley. The 
directions of the rotation of the shaft D and therefore of the 
movement of the machine table are thus reversed until the table 
reaches the other end on its stroke whereupon the parts are 
returned to the position shown and the cycle of operations is 
repeated. 

The fast pulley is often made narrower than the idle pulleys 
and the distance apart of the belt forks arranged so that in an 
intermediate position of the latter each belt is on an idle pulley ; 
in this way the operation of the machine can be stopped when 
required. 

Belt Drive for Inclined Axes. Up to the present we have 
considered only the case in which the axes of the shafts to be 
driven by belting are parallel. 

Fig. 173 shows a way of providing for two axes at right 
angles to each other provided that they are not too close. This 
will drive satisfactorily in the direction shown ; the arrangement 



XVII] BELT, CHAIN, AND TOOTHED GEARING 279 

must be such that the middle point of the width of the belt 
where it leaves one pulley is in the central plane of the other 
pulley. 




Fig. 173. Belt Drive for Inclined Axes. 

Where the position of the shafts is such that a direct drive 
cannot be effected, guide pulleys must be used. Fig. 174 shows 
one such arrangement, Q^ G being the guide pulleys. 




Fig. 174. Belt Drive with Guide Pulleys. 

Toothed Gearing. Suppose that two smooth discs X, Y 
(Fig. 175) rotate in contact without slipping. 

Then in one revolution of the driver X a point on the circum- 
ference moves through a distance ttDx. If therefore there is no 
sKp between the two discs, a point on the circumference of the 
follower Y must move through the same distance ttDx* But 
one revolution of the follower corresponds to irDy, so that the 
number of revolutions of the follower for one of the driver 

ttDy Dy' 

number of revs, of follower , ., ,. Dx 

:. r 7 j-r-' = velocity ratio = r^... .(1). 

number of revs, of driver *' Dy ^ ' 

Now suppose that in order to prevent any possibility of 



280 



BELT, CHAIN, AND TOOTHED GEARING [ch. 



slipping we form teeth upon the surfaces of these discs. For 
simplicity in the figure only a few teeth are shown, but it will 
be understood that they are formed all round the wheel. 

It is clear that these teeth must be of special shape if they 
are to mesh and roll into action smoothly. The curve most 
commonly used for gear teeth is called the involute. 



Pitch circles 




Fig. 176. Toothed Gearing. 

The form of toothed gear shown in Fig. 175 is called spur 
gearing. The axes of the two shafts are parallel and the teeth 
are straight and usually run at right angles to the plane of the 
wheels. 

The circles which correspond to the untoothed or smooth discs 
are called the pitch circles. The distance upon the pitch circle 
between the centres of two succeeding teeth is called the pitch, 
or more accurately the circular pitch p. It is equal to the 
circumference of the pitch circle divided by the number of teeth. 
The diametral pitch m is equal to the diameter of the pitch circle 
divided by the number of teeth and can be obtained by dividing 
the circular pitch ^ by tt (3-1416). The diametral pitch is 
sometimes called the module. Other forms of toothed gearing 
are shown in Figs. 176-179. 



xvn] BELT, CHAIN, AND TOOTHED GEARING 281 

In the rack and pinion, Fig. 176, one of the members is straight, 
this corresponding to spur gearing in which one of the wheels is 



Pinion 




Fig. 176. Rack and Pinion, 
infinitely large. A rotation of the pinion causes a rectilinear 
movement of the rack. 

Bevel gearing, Fig. 177, is used to connect two shafts at an 




Fig. 177. Bevel Gearing, 
angle to each other (usually a right angle) and meeting at a point. 
It corresponds to a toothed form of two smooth cones rotating in 
contact. 



282 BELT, CHAIN, AND TOOTHED GEARING [CK. 




Fig. 178. Spiral Gearing. 




^'ig. 179. Worm Gearing, 



XVII] BELT, CHAIN, AND TOOTHED GEARING 283 

Spiral gearing, Fig. 178, is used to connect two shafts at an 
angle to each other which do not meet at a point. 

In worm gearing, Fig. 179, the shafts are at right angles to 
each other and do not intersect. The driver X is a worm or 
screw and the follower F is a wheel whose teeth are formed to 
gear accurately with the worm. The velocity ratio in such gears 
is small and as a rule the gear cannot be reversed, i.e. the wheel 
cannot drive the worm. As was shown on p. 66 this means that 
the efficiency of the gear cannot be greater than 50 %, but in 
many machines this objection is of minor importance compared 
with the advantage that the gear is self-locking, i.e. that it will 
not run backwards if the drive is removed. 

Velocity Ratio in Toothed Gear Trains. The term "gear 
train" is used to indicate a number of gear wheels working in 
combination. 

For a pair of spur wheels we have seen that 

, ., , . diameter of driver 

velocity ratio = v- = 3-^ 7 . . ,, . 

'' ^ diameter of follower 

Now the circular pitch of the two wheels must be the same 
and the number of teeth x pitch must be equal to the circum- 
ference of the pitch circle. 

.*. we have 

nxp = TrDjr (2), 

and UyP = ttD^ (3). 

Dividing we get 

.*. from (1) velocity ratio = v^ = "^ (5). 

Expressed in words : 

... . _ number of teeth on driver 
number of teeth on follower ' 
number of revs, of follower number of teeth on driver 
number of revs, of driver number of teeth on follower * 

i.e. ^ = !^ = ^ (6). 

Nx ny Dy 

Numerical Examples. (1) A toothed wheel of 10 inches 
diameter on the pitch line and with 60 teeth runs at 120 revolutions 
ver minute and drives a wheel of 4 inches diameter o 



284 BELT, CHAIN, AND TOOTHED GEARING [ch. 

Find (a) the circular pitch of the teeth, (6) the diametral pitchy 
(c) the nvimher of teeth on the second wheels {d) the number of 
revolutions which it will make. 

(a) np = ttD, 

/. 60^} = 3- 1416 X 10, 
^ 31-416 

= '524 inch . 

,, , TA- ^ 1 -J. 1- diam. of wheel 

(o) Diametral pitch = m = r r-; — ti- 

^ ' ^ number of teeth 

_ 1 

= -167 inch , 
(c) From result (4) we have 

ny Dy' 

60^10 
ny~ 4 ' 
.'. lOny = 240, 
/. ny = 24. 

{d) In this case v, = |f from (5) 

= 2-5. 
^ _ revolutions of follower 

*" revolutions of driver ' 

/. revolutions of follower = 120 x 2-5 = 300 per minute. 

(2) Toothed wheels of 2} inches pitch are required to connect two 
shafts running at 340 and 115 revolutions per minute , the centres of 
the wheels to be as nearly as possible Sft, apart. Find suitable 
numbers of teeth for the wheels. 

The distance apart of the centres of two toothed wheels is 
equal to the sum of the radii of the pitch circles, i.e. equal to 
half the sum of the diameters of the pitch circles. 



i.e. distance apart 



Dx-^Dy 



2 
Now we have ^ = ^ , 

Nx Dy 

^_340^68 
'' Dy 115 ~ 23* 
Dx±Dy^6S + 23 91 
Dx ~ 23 ^23* 



XVII] BELT, CHAIN, AND TOOTHED GEARING 285 



Now take 


Dx-h Dr = 6 ft. = 6 X 12 inches, 




^ 6x12x23 ,oo- u 
•'• ^x= oi = 1^'^ inches. 




/. Dy = 72 - 18-2 = 53-8 inches. 


Now 


rixP = TrDxf 


and 


riyp = ttDy, 




. TT X 18-2 ,^^ 



.•.^z= -^-^7^^-22 9, 

TT X 53-8 
Wf= — ^Tg — = b7b. 

But the number of teeth must be a whole number, 
.*. take Ux = 23 and Wj. = 68. 

These will give the required velocity ratio. The student 
should note carefully that in problems of this kind it is essential 
that the numbers of teeth be chosen to give the exact velocity 
ratio required. 

Idle Gear Wheels. In the use of spur gearing it is often 
necessary to use wheels intermediate between the driver and the 
follower, as shown in Fig. 180, such wheels being called "idle 




Fig. 180. Idle Gear Wheels. 

wheels " or " idlers. " These idle wheels are used either to reverse 
the direction of rotation or else to enable the distance between 
the two shafts to be greater than the sum of the radii of the 
driver and follower. 

Idle wheels have no effect on the velocity ratio. If the number 
of idle wheels is odd the driver and follower rotate in the same 
direction, but if even they rotate in opposite directions. 

Compound Gear Trains. To obtain a larger or smaller velocity 
ratio than is practicable with one pair of spur wheels, compound 



286 BELT, CHAIN, AND TOOTHED GEARING [ch. 



gear trains such as shown in Fig. 181 are usually employed. 
Such a gear train will, for instance, be found in every watch or 
clock, the form shown giving a small velocity ratio. 

The wheel A gears with a wheel B which is formed solid with 
or is keyed to the same shaft as a wheel C; tLis drives a spur 
gear D which is coaxial with a wheel E which gears with the 
follower F. It is usual to refer to the alternate wheels A, Cy E 
as drivers and the wheels B, D, F s& followers. 

We wiU trace out the compound velocity in steps Ua, ns etc., 
being the number of teeth ia the various wheels, and iV^, N^ etc., 
their number of revolutions per minute, it being noted that N^ 
must be equal to Nq and iV^ equal to iV^. 




Fig. 181. Compound Gear Train. 
Considering the first pair of wheels Ay B, we have 



N'a 



Ub 



:. Nb = Na.^ .... 

Ub 

:, since Nc = Nb, 
Considering the wheels C and D, we have 

nc 



(7), 
(8). 



= Na 



rir 



. ^ (from 8) 
Ub n^^ 



:. since N j^ = N^y 

N^^Na.'^.'^ 
Ub Uii 



.(9), 
(10). 



XVII] BELT, CHAIN, AND TOOTHED GEARING 287 



I.e. 



Considering the wheels E and J^, 


we have 






Np Ue 
Ne np* 








:. Np = Ne 


Tip 






^N^ 




'rip 


. Np 


velocity of compound 1 


brain 




n^ xrio X fig 








n^ X Tij) X Up 




product 


, of number of teeth in 


drivers 



(11). 



....(12). 



product of number of teeth in followers 
This formula can be used for a compound train of any number 
of pairs. 




Lathe 
Spindle 



Fig. 182. Back Gear Drive of a Lathe Spindle. 

Sometimes, as in the back gear drive of a lathe spindle, 
Fig. 182, the driver and the follower are arranged on the same 
shaft, one of them being loosely mounted. In the normal working 
of the lathe the back shaft is moved backwards a little to bring 
the wheels out of gear and the cone pulley, which runs loosely 
upon the lathe spindle and has the wheel A integrally connected 
to it, is connected by a radially movable pin Oy which enters 
between projections F on the pulley, the wheel D being keyed 
to the lathe spindle. To bring the back shaft into operation the 
pin O is released and the back gear is brought into engagement as 
shown. The same rule is used for the velocity ratio, no matter 
how the shafts are arranged. 



288 BELT, CHAIN, AND TOOTHED GEARING [ch. 

Numerical Examples on Compound Trains. 
(1) In a compound gear train the driver A has 40 teeth and 
gears with a wheel B with 20 teeth. Keyed on the same shaft as 
B is a wheel C of 120 teeth gearing with the follower D with 75 
teeth. If the wheel A runs at 35 revolutions per minute, how many 
revolutions per minute will the wheel D make ? 
In this case 

_ product of teeth in drivers 
' product of teeth in followers 
40 X 120 



20 X 75 

_48 
~I5' 
number of revolutions of follower _ 48 
number of revolutions of driver 15 ' 

• * 35 ~ 15 ' 
,.^,=4^^ = 112. 

(2) In a lathe headstock the lowest direct drive is 50 revolutions 
per minute and the back gear has to be designed so as to reduce this 
to 5 revolutions per minute. Find suitable numbers of teeth for the 
various wheels of the back gear. 

In this case we see that as the wheels A,B and C, D each form 
a pair whose axes are at the same distance apart, the sum of the 
radii of each pair of wheels must be the same, that is to say the 
sum of the number of teeth must be the same for each pair, if 
the teeth have the same pitch. In this case 

5 1 



we have 



^^ 50 10 * 
1 n^x nc 



10 nB X nj) 
Further n^ + Wj, = w^ + nj). 

Suppose n^ = 20. 

Then if we take n^,. = 40, nc = 10, and w^ = 50, this gives us 

20 X 10 1 



v« = 



40x50 10" 



XVII] BELT, CHAIN, AND TOOTHED GEARING 289 

Reversing Tooth Drive for Lathe Lead Screw. The following 
arrangement of gearing is commonly employed for driving the 
lead screw L (Fig. 183) of a lathe from the headstock spindle 0. 
The headstock spindle carries a toothed wheel A which drives 
a pinion D mounted on a spindle X either directly through a 
pinion B or else through a pinion C. The pinions B, C are moimted 
in a plate E pivoted on the shaft of the pinion D and provided with 
a slot engaging a stop-pin F for fixing it in its extreme positions. 
In the position {a) shown in the figure the pinion D is rotated in 
O 




Kg. 183. Reversing Drive for Lathe Lead Screw. 

the same direction as the pinion A whereas in the position (6) the 
pinion D is driven in the opposite direction to the pinion A. 
In this position the pinion B has gone out of contact with the 
pinion A and the pinion C has come into contact with it. From 
the spindle X the drive goes through change wheels P, Gy H and 
J adjustably carried in an arm K. The wheel H engages a wheel 
J on the lead screw shaft. The sizes of the wheels G, H and J 
are so chosen as to give the required velocity ratio. The spindle 
of the pinions G, H is adjusted in the slot in the quadrant so that 
the wheel H meshes correctly with the wheel J and the quadrant 
is then adjusted by means of a curved slot M so as to bring the 
wheel G into correct mesh with the wheel P. The quadrant is 
kept in its adjusted position by means of a locking bolt N. 
&.. M. 19 



290 BELT, CHAIN, AND TOOTHED GEARING [en. 



The number of teeth on the wheel D is usually equal to that 
on the wheel A so that the spindle X rotates at the same speed 
as the headstock spindle. 

Numerical Example. The leading screw of a lathe is f pitch 
and it is required to cut a screw of 10 threads per inch. Find 
suitable sizes of the gear wheels. 

For one revolution of the lead screw the lathe saddle will be 
moved | in., but for one revolution of the lathe spindle we wish 
the saddle to be moved onJy ^in. 

Now the saddle will be moved 1 in. in f revolutions and there- 
fore will be moved ^ in. in -^q revolutions so that we have to 
choose our change gears P, G, H and J so as to give a velocity 
ratio of ^ = f^. 

The lathe is provided with a whole set of wheels of different 
numbers of teeth usually rising five at a time. 

Suppose we take Up = 30, Uq = 75, n^ = 30 and Uj = 90. 

np.ns_ 30x30 _ 2_ 
15' 



This will give 



v^ = 



Uq.Uj 

and this is the ratio required. 



75x90 




Bevel Gear Reversing Train. The following arrangement of 
bevel gearing is commonly adopted as a convenient reversing 
mechanism for a shaft. The 
drive goes from the shaft A 
which has feathered thereto a 
double clutch jaw F. A bevel 
wheel C is loosely mounted on 
the shaft A and engages a bevel 
wheel D, the other end of which 
engages a bevel wheel E fixed to a 
shaft B in line with the shaft A . 

The bevel wheels E and C each 
carry clutch jaws for engagement with the jaws on the clutch jaw 
F. In the position shown the rotation of the shaft A is trans- 
mitted direct from the clutch jaw to the bevel wheel E and thus 
to the shaft B, the wheels D and C rotating idly. The shafts 
A and B then rotate in the same direction. If the clutch F is 
moved to the right so as to engage with the wheel C the drive 
goes through the wheels C, D and E to the shaft B which then 
rotates in an opposite direction to the shaft A. 



Fig. 184. Bevel Gear Reversing 
Train. 



xvn] BELT, CHAIN, AND TOOTHED GEARING 291 



SUMMARY OF CHAPTER XVn. 

Belt Gearing, h.p. transmitted = ^33000 • 

Ti may be taken as 2Tj in the absence of more exact information. 

,_ - . , , . diameter of driver 

Velocity ratio =37 titti • 

•^ diameter of follower 

In " open " belt the driver and follower rotate in the same 
direction, and in "crossed" belt they rotate in opposite directions. 

Cone Pulleys. Open belts. If S is the distance apart of the shafts, 

the quantity ^ (Z)^+Dy) + — ^ .^ must be constant. 

Crossed belts. The sum of the diameters of corresponding 
pTilleys must be constant. 

Toothed Gearing. 

T^. ^ 1 .^ 1. circular pitch diameter of pitch circle 

Diametral pitch = = ^ ^-^- — -i . 

^ TT number of teeth 

^. , .^ , circumference of pitch circle 

Circular pitch = r , f ^, . 

^ number of teeth 

,_ - . . number of teeth on driver 

Velocity ratio = r 7-7 — re: lnr^ • 

•^ number of teeth on follower 

Velocity ratio of compound train 

product of number of teeth on drivers 



product of number of teeth on followers * 

Idle wheels only alter the direction of rotation; they do not 
affect the velocity ratio. 



EXERCISES. XVII. 

1. A shaft is to be driven at 400 revolutions per min. and carries 
a pulley of 8 ins. diameter. What size driving pulley is necessary 
for a shaft which has to be driven from it at 70 revolutions per 
minute ? 

2. Two shafts at right angles to each other have to be driven 
by bevel gearing, the driving shaft runs at 120 revolutions per min. 
and carries a wheel with 48 teeth on it. How many teeth must 
be placed upon the second wheel if its shaft has to run at 320 revolu- 
tions per minute? 

19—2 



292 BELT, CHAIN, AND TOOTHED GEARING [ch. 

3. If a belt transmits 25 h.p. at 150 revolutions per minute 
over a pulley 3 ft. diameter find the difference of tension on the 
two sides of the belt. 

If the tension on the tight side is three times that on the slack side 
find the tension on each side. 

4. The crank of an engine is 2 ft. in length, and the diameter 
of the flywheel is 10 ft., also the flywheel has teeth on its rim 
and drives a pinion 3 ft. in diameter. If the mean pressure on the 
cranio pin is 7^ tons, what is the mean driving pressure on the teeth 
of the pinion? 

5. A friction wheel 4 ft. diameter running at 70 revs, per min. 
drives a wheel 2 ft. 3 ins. diameter. Find the force with which 
the wheels must be pressed together per h.p. transmitted when 
the coefficient of friction for the surfaces is '15. 

6. In a lifting crab the length of the handle is 1 6 ins. and diameter 
of barrel 8 ins. The pinion on the same axis as the handle has 16 
teeth, and gears with the spur wheel connected to the barrel which 
has 90 teeth. What weight can one man exerting a push of 30 lbs. 
lift? 

7. The preceding is fitted to act with an increased velocity 
ratio by sliding the pinion out of contact with the spur wheel, and 
putting in gear a pinion of 18 teeth working with a spur wheel of 
54 teeth. On the axis of the latter is another pinion of 18 teeth 
which now drives the 90 wheel. Find the force required to lift 
1 ton. 

8. The annexed sketch (Fig. XVII a) shows the arrangement of 
pulleys and belts used for driving a dynamo machine F from the 
steam-engine A. 

Diameter of ^ = 57", B = 36", 
„ C = 42^ D = 24", 
,y E = 48", F = 14". 




f 



Fig. XVII a. 

If the speed of A is 96 revs, per min. find the speed of F, assuming 
there is no slipping of belts. 



XVII] BELT, CHAIN, AND TOOTHED GEARING 293 

9. A machine is driven from a pulley 4 ft. in diameter by means 
of a belt. If the difference of pull in the two sides of the belt is 
20 lbs. weight, and the pulley miakes 120 revolutions per min., find 
the H.P. transmitted by the belt. 

1 0. The saddle of a lathe weighs 5 cwt. ; it is moved along the 
bed by a rack and pinion arrangement. What force applied at 
the end of a handle 10'' long will be capable of just moving the saddle, 
supposing the pinion to have 12 teeth of IJ* pitch and the coefficient 
of friction between the saddle and the lathe bed to be • 1, other friction 
being neglected? 

11. A leather belt J inch thick has to transmit 10 h.p. from 
a pulley 4 ft. in diam. making 120 revolutions per minute. Assuming 
that the tension on the tight side is twice that on the slack side 
find the width of belt necessary if the safe stress in the belt is 
320 lbs. per sq. in. 

12. The tension per inch width of a belt must not exceed 1 10 lbs. 
Find the width required to transmit 12 h.p. from a shaft running 
at 80 revolutions per minute. 

D = 4:it. 6 ins., ^' = 1|. 

13. A pulley 4 ft. in diameter is driven by two belts running 
over each other, each f in. thick. The speed of the middle plane 
of the inner belt is 1800 ft. per minute. How much does the outer 
gain on the inner per minute? 

14. The set of wheels for a screw cutting lathe range from 
20 to 150 teeth, there being two 20 wheels. The leading screw has two 
threads to the inch. Arrange suitable trains for cutting threads 
on a I in. screw, 20 threads to the inch. 

15. The greatest and least diameters of the pulleys of a speed- 
cone for a headstock mandrel are 10* and 5 J" respectively ; and this 
speed-cone is driven from a similar speed-cone keyed to a counter- 
shaft which makes 250 turns per min. The back gearing is of the 
usual type, the spur wheels concentric with the headstock spindle 
having 62 and 30 teeth gearing with wheels having 18 and 50 teeth 
respectively on the back spindle. Find the greatest and least 
revolutions per min. at which the headstock mandrel ma3'' be driven. 

16. The effective diameter of a worm is 6* and the pitch of 
the thread of the worm 2V. The worm is secured on the shaft 
of an engine of 60 b.h.p. and gears with a wheel on a shaft whose 
axis is at right angles to that of the engine shaft. li fx = -IQ find 
^ and H.p. transmitted by second shaft. 



APPENDIX 

THE SUM CURVE CONSTRUCTION 

The sum curve can be obtained graphically as follows. Let 
ACD, Fig. a, be any primitive curve on a straight base AB. 
Divide AB into any number of parts, not necessarily equal (but 
for convenience of working they are generally taken as equal). 




' < 34 S 6 7«9lO|lA 

Fig. o. 



These so-called base elements should be taken so small that the 
portion of the curve above them may be taken as a straight line. 
About 1 cm. or '4 in. will usually be a suitable size and in most 
cases V, smaller element, 11, will come at the end. Find the mid- 
poi^if^, 1, 2, 3, etc., of each of the base elements and let the 



THE SUM CURVE CONSTRUCTION 295 

verticals through these mid -points meet the curve in la, 2a, 3a, 
etc. Now project the points on to a vertical line AE, thus 
obtaining the points 16, 26, 36, etc., and join such points to a 
pole P on AB produced and at some convenient distance p from 
A. Across space 1 then draw Ad parallel to P16, de across 
space 2 parallel to P26, and so on, until the point n is reached. 
Then the curve Ade. , .n is the sum curve of the given curve, 
and to some scale Bn represents the area of the whole 
curve, 

Peoof. Consider one of the elements, say 4, and draw fo 
horizontally. 

Now A/gro is similar to the A P4:bA, 

. go_4:b,A 
" fo~ PA ' 

but PA = p and 46, A = 4, 4:a, 

fox 4:, 4:a area of element 4 of curve 
.*. go ='' — = . 

„, ., I - area of element 3 of curve , 
Similarly /g = and so on, 

.*. ordinate through gr = gro + /g + . . . 
_ area of first four elements of curve 

.*, the curve Ade. . .w is the sum curve required. 

Then if Bn be measured on the vertical scale and p be measured 
on the horizontal scale, the area of the whole cuTTe will be equal 
to p X Bn. 

It is obviously advisable to make p some •onvcniont round 
number of units. 

The sum curve obtained by this method may hare the same 
operation performed on it, and thus the second sum curve of the 
primitive curve is obtained, and so on. 

If the operation be performed on a rectangle, the sum curve 
will obviously become a sloping straight line, and if the sum 
curve of a slopmg straight line be drawn, it will be found to 
be a parabola. In the case in which it is required to apply this 
construction to a curve which is not on a straight base, the curve 
is first brought to a straight base as follows ; 



296 



APPENDIX 



Suppose AcBdy Fig. 6, is a closed curve. Draw verticals 
through AB to meet a horizontal base A'B'. Divide the curve 
into a number of segments by vertical lines at short distances 
apart, and set up from the base A'B' lengths a^, 61, etc., equal to 




Fig. b. 



the vertical portions a, b, etc., on the curve. Joining up the 
points thus obtained we get the corresponding curve A'c^B' on 
a straight base. 



TRIGONOMETRICAL RELATIONS 



297 



SID A==- 
o 



RIGHT-ANGLED TRIANGLES 
h 



qobA 



BQC A=-- 



cosec4=- 
a 



tan4=:- 



cotan A = 




tan 6 = 



sin 



COS0 

sin'^ + cos«^ = l 
l+tan2^=sec2 
versine ^ = 1 - cos 9 
coversine = 1 - sin ^ 
8in(^ + 0)=sin 6 cos^+cos ^ sin 



Complement of ^= 90° - 
Supplement of ^ = 180° - 6 

cos(^ + 0)=cos ^cos0 -sin 6 sin 
tan ^+tan <p 



tan (0+0) = , , 

^ ^ l-tan0tan0 

sin 20 =2 sin 6 cosd 

cos20=cos2 0-sin2tf 

2tan0 



tan 26 = 



l-tan2 



Given 


Required 


Formulae 




AyC,C 

A,C,b 

C,c,b 
C,a,c 
C,a,b 


A « 

sm A =r 



ta,nA=- 
c 

C=90°-^ 

C=90°-^ 

C=90°-A 


COsC = r 


cotanC=- 
c 

c=a Xcotan^ 

a=6xsini4 

a = cX tan ^ 




a, 6 


c=\/(6+o)(6-a) 


a,c 

A, a 
A,b 
A,c 


6=N/(a2+c2) 

smA 
c=bx cos A 

b- ^ 


coSi4 



OBLIQUE-ANGLED TRIANGLES 



Given 



A, B, C, a 
A,b,c 
a, b, c 



Formulae 



(o^ X sin £ X sin C)-^2 sin A 
Area=< |(cx6xsin4) 



J 



( \/s {s -a){s- b) (.• - c) 





Formulae 



c = a 



sin C = 



tani4 = 



sin C 

sin^ 

csin4 

a 

a sin B 
c -a cos B 



a, b, c 



sm 4^ = \/ ^ =-^-^ - 

. . /s{8-a) 

C08\A= ^ 



bxc 



; tan 



IA=^' 



{8 -b){s-c) 

9{s -a) 



298 



LOGARITHMS. 



10 





1 


2 


3 


4 


5 


6 


7 


8 


9 


1 2 


3 


4 5 


6 


7 8 9 


0000 


0043 


0086 


012S 


0170 


0212 


0253 


0294 


0334 


0374 


4 8 


12 


17 


21 


25 


29 33 37 


ZI 
12 

13 


•0414 
•0792 

•1139 


0453 
0S28 

1173 


0492 
0864 
1206 


0531 
0899 
1239 


0569 

0934 
1271 


0607 
0969 

1303 


0645 
1004 

1335 


0682 
1038 
1367 


0719 
1072 
1399 


0755 
1 106 

1430 


4 8 
3 7 
3 6 


II 

10 
10 


15 
14 
13 


19 23 
17 21 
16 19 


26 30 34 
24 28 31 
23 26 29 


14 
15 
i6 


1461 
•I76I 
2041 


1492 
1790 
2068 


1523 
1818 
2095 


1553 
1847 
2122 


1584 
1875 
2148 


1614 

1903 

2175 


1644 

1931 
2201 


1673 

1959 
2227 


1703 
1987 
2253 


1732 

2014 
2279 


3 6 
3 5 


1 

8 


12 
II 
II 


15 
14 
13 


18 

17 
16 


21 24 27 
20 22 25 
18 21 24 


H 
x8 

19 


•2304 

•2553 
•278S 


2330 

2577 
2810 


2355 
2601 

2833 


2380 
2625 
2856 


2405 
2648 
2878 


2430 
2672 
2900 


2455 
2695 
2923 


2480 
2718 
2945 


2504 

2742 
2967 


2529 
2765 
2989 


2 5 
2 5 
2 4 


7 
7 
7 


10 
9 
9 


12 

12 
II 


15 

14 
13 


17 20 22 
16 19 21 

16 18 20 


20 


•3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2 4 


6 


8 


II 


13 


15 17 19 


21 
22 
23 


•3222 

•3424 
•3617 


3243 
3444 
3636 


3263 

3464 
365s 


3284 
3483 
3674 


3304 
3502 
3692 


3324 
3522 

3711 


3345 
3541 
3729 


3365 
3560 

3747 


3385 

3579 
3766 


3404 
3598 
3784 


2 4 

2 4 
2 4 


6 
6 
6 


8 
8 
7 


10 

10 

9 


12 
12 
II 


14 16 iS 
14 15 17 
13 15 17 


24 

26 


•3802 

•3979 
•4150 


3820 

3997 
4166 


3838 
4014 

4183 


3856 
4031 
4200 


3874 
4048 
4216 


3892 
4065 
4232 


3909 
4082 

4249 


3927 
4099 

4265 


3945 
4116 
4281 


3962 

4133 
4298 


2 4 
2 3 
2 3 


5 
5 
5 


7 
7 
7 


9 
9 

8 


II 

10 
10 


12 14 16 
12 14 15 
" 13 15 


28 
29 


•4314 

•4472 

•4624 


4330 

4487 

4639 


4346 
4502 
4654 


4362 
4518 
4669 


4378 

4533 
4683 


4393 
4548 
4698 


4409 
4564 
4713 


4425 
4579 
4728 


4440 

4594 
4742 


4456 
4609 

4757 


2 3 
2 3 
I 3 


5 
5 
4 


6 
6 
6 


8 
8 
7 


9 
9 
9 


II 13 14 
II 12 14 
10 12 13 


30 


•4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


I 3 


4 


6 


7 


9 


10 II 13 


31 
32 
33 


•4914 
•5051 
•5185 


4928 
5065 
5198 


4942 

5079 
521 1 


4955 
5092 

5224 


4969 
5105 
5237 


4983 
5119 

5250 


4997 
5132 
5263 


501 1 

5145 
5276 


5024 

5159 
5289 


503S 
5172 
5302 


I 3 
I 3 
I 3 


4 
4 
4 


6 
5 
5 


7 
7 
6 


8 
8 
8 


10 II 12 
9 II 12 
9 10 12 


34 
36 


•5315 

•5441 
•5563 


5328 
5453 
5575 


5340 
5465 
5587 


5353 
5478 
5599 


5366 
5490 
561 1 


5378 
5502 
5623 


5391 
5514 
5635 


5403 
5527 
5647 


5416 

5539 
5658 


5428 
5670 


I 3 
I 2 
I 2 


4 
4 
4 


5 
5 
5 


6 
6 
6 


8 


9 10 II 

9 10 II 
8 10 II 


37 
38 
39 


•5682 

■579« 
*S9" 


5694 

5809 

5922 


5705 
5821 

5933 


5717 
5832 
5944 


5729 
5843 
5955 


5740 

5855 
5966 


5752 
5866 

5977 


5763 
5877 
5988 


5775 

5888 

5999 


5786 
5899 
6010 


I 2 
I 2 
I 2 


3 
3 
3 


5 
5 
4 


6 
6 
5 




8 9 10 
8 9 10 
8 9 10 


40 


•6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


I 2 


3 


4 


5 


6 


8 9 10 


41 
42 

43 


•6128 
•6232 

•6335 


6138 
6243 

6345 


6149 
6253 
6355 


6160 
6263 
6365 


6170 

6274 

6375 


6180 
6284 
6385 


6191 

6294 


6201 
6304 
6405 


6212 
6314 
6415 


6222 

6325 
6425 


I 2 
I 2 
I 2 


3 
3 
3 


4 
4 

4 


5 
5 
5 


6 
6 
6 


7 8 9 
7 8 9 

789 


44 
46 


'6435 
•6532 
•6628 


6444 
6542 
6637 


6454 
6551 
6646 


6464 
6561 
6656 


6474 

6571 
6665 


6484 
6580 
6675 


6493 
6590 
6684 


6503 
6599 
6693 


6513 
6609 

6702 


6522 
6618 
6712 


I 2 

I 2 
I 2 


3 
3 
3 


4 

4 
4 


5 

5 
5 


6 
6 
6 


7 8 9 
7 8 9 
7 7 8 


47 
48 

49 


•6721 
•6812 
•6902 


6730 
6821 
6911 


6739 
6830 
6920 


6749 

6839 
6928 


6758 
6848 
6937 


6767 
6857 
6946 


6776 

6866 
6955 


6785 

6875 
6964 


6794 

6884 
6972 


6803 

6893 
6981 


I 2 
I 2 
I 2 


3 
3 
3 


4 
4 
4 


5 
4 
4 


5 
5 
5 


678 

678 
678 


50 


•6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


I 2 


3 


3 


4 


5 


678 


51 
52 
53 
54 


•7076 
■7160 

•7243 
•7324 


7084 
7168 
7251 
7332 


7093 
7177 

7259 
7340 


7101 
7185 
7267 
7348 


7110 
7193 
7275 
7356 


7118 
7202 
7284 
7364 


7126 
7210 
7292 
7372 


7135 
7218 
7300 
7380 


7143 
7226 

7308 

7388 


7152 

7235 
7316 
7396 


I 2 
I 2 
I 2 
I 2 


3 
2 
2 

2 


3 
3 
3 
3 


4 

4 
4 

4 


5 
5 
5 
5 


678 
677 
667 
667 



LOGARITHMS. 



299 








1 


2 


3 


4 


s 


6 


7 


8 


9 

7474 
7551 


1 2 3 


4 


6 6 


— " 
7 S 9 


56 


7404 

7482 


7412 
7490 


7419 

7497 


7427 
7505 


7435 
7513 


7443 
7520 


7451 
752S 


7459 
7536 


7466 

7543 


122 
I 2 2 


3 
3 


4 5 
4 5 


5 t 7 
5 6 7 


59 


•7559 
•7634 
7709 


7566 
7642 
7716 


7574 

7649 

7723 


7582 
7657 
7731 


7589 
7664 

7738 


7597 
7672 

7745 


7604 
7679 
7752 


7612 
7686 
7760 


7619 

7694 
7767 


7627 

7701 
7774 


I 2 2 
112 
I I 2 


3 
3 
3 


4 5 
4 4 
4 4 


5 ^ 7 
5 6 7 
5 6 7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


112 


3 


4 4 


566 


61 
62 
63 


•7853 

7924 
•7993 


7860 
7931 

8000 


7868 

7938 
8007 


7875 
7945 
8014 


7882 
7952 
S021 


7889 

7959 
8028 


7896 
7966 

8035 


7903 
7973 
8041 


7910 
7980 
8048 


7917 
7987 
8055 


112 
112 
112 


3 
3 
3 


4 4 
3 4 
3 4 


566 
566 
5 5 6 


65 
66 


•8062 
•8129 

•8195 


8069 
8136 

8202 


8075 
8142 

8209 


8082 
8149 
8215 


8089 
8156 
8222 


8096 
8162 
8228 


8102 
8169 
8235 


8109 
8176 

8241 


8116 

8182 
8248 


8122 
8189 
8254 


I I 2 
I I 2 
112 


3 
3 
3 


3 4 
3 4 
3 4 


5 5 f 
5 5 6 
5 5 6 


69 


•8261 

•8325 

•S388 


8267 
8331 

8395 


8274 

8338 
8401 


8280 

8344 
8407 


8287 

8351 
8414 


8293 
8357 
8420 


8299 

8363 
8426 


8306 
8370 
8432 


8312 

8376 
8439 


8319 
8382 

8445 


112 

112 
112 


3 
3 
2 


3 4 
3 4 
3 4 


5 5 6 
456 
456 


70 


•8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


I I 2 


2 


3 4 


456 


71 
72 

73 


•8513 
•8573 
•8633 


8519 
8579 
8639 


8525 

8585 
8645 


8531 
8591 
8651 


8537 
8597 
8657 


8543 
8603 
8663 


8549 
8609 
8669 


8555 
8615 

8675 


8561 
8621 
8681 


8567 
8627 
8686 


112 
112 
112 


2 

2 
2 


3 4 
3 4 
3 4 


4 5 5 
4 5 5 
4 5 5 


74 


•8692 

•8751 
•8808 


8698 

8756 

8814 


8704 
8762 
8820 


8710 

8768 
8825 


8716 

8774 
8831 


8722 
8779 
8837 


8727 
8785 
8842 


8733 
8791 
8848 


8739 
8797 
8854 


8745 
8802 

8859 


112 
112 
I I 2 


2 
2 
2 


3 4 
3 3 
3 3 


4 5 5 
4 5 5 
4 5 5 


79 


•8865 

■8921 

•8976 


8871 
8927 
8982 


8876 
8932 
8987 


8882 
8938 
8993 


8887 

8943 
8998 


8893 
8949 
9004 


8899 

8954 
9009 


8904 
8960 
9015 


8910 
8965 
9020 


8915 
8971 
9025 


112 
I I 2 
112 


2 
2 
2 


3 3 
3 3 
3 3 


4 4 5 
4 4 5 
4 4 5 


SO 


•9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


I I 2 


2 


3 3 


4 4 5 


8x 
82 
83 


•9085 

•9138 
•9I9I 


9090 

9143 
9196 


9096 
9149 
9201 


9101 

9154 
9206 


9106 

9159 
9212 


9112 
9165 
9217 


9117 
9170 
9222 


9122 

9175 
9227 


9128 
9180 
9232 


9133 
9186 

9238 


I I 2 
112 
112 


2 
2 
2 


3 3 
3 3 
3 3 


4 4 5 
4 4 5 

4 4 5 


?4 
85 

86 


•9243 

•9294 

•9345 


9248 
9299 
9350 


9253 
9304 
9355 


9258 

9309 
9360 


9263 
931S 
9365 


9269 
9320 
9370 


9274 
9325 
9375 


9279 

9330 
9380 


9284 
9335 
9385 


9289 
9340 
9390 


I I 2 
112 
I I I 


2 
2 
2 


3 3 
3 3 
3 3 


4 4 5 
4 4 5 
4 4 5 


87 
88 

89 


•9395 

•9445 
•9494 


9400 
9450 

9499 


9405 

9455 
9504 


9410 
9460 
9509 


9415 
9465 
9513 


9420 
9469 
9518 


9425 
9474 
9523 


9430 

9479 
9528 


9435 

9484 

9533 


9440 
9489 
9538 


I I 
I I 
I I 


2 
2 
2 


2 3 
2 3 
2 3 


3 4 4 
3 4 4 
3 4 4 


90 


•9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


I I 


2 


2 3 


3 4 4 


91 
92 

93 


•9590 

•9638 
•9685 


9595 

9643 
9689 


9600 
9647 
9694 


9605 
9652 
9699 


9609 
9657 
9703 


9614 
9661 
9708 


9619 
9666 

9713 


9624 
9671 
9717 


9628 

9675 
9722 


9633 
9680 

9727 


I I 
I 1 
oil 


2 
2 
2 


2 3 
2 3 

2 3 


3 4 4 
3 4 4 

3 4 4 


94 
96 


•9731 

•9777 

•9823 


9736 
9782 
9827 


9741 
9786 

9832 


9745 
9791 
9836 


9750 

9795 
9841 


9754 
9800 

9845 


9759 
9805 
9850 


9763 
9809 
9854 


9768 
9814 
9859 


9773 
9818 
9863 


oil 
I I 
oil 


2 
2 
2 


2 3 
2 3 
2 3 


3 4 4 
3 4 4 
3 4 4 


97 -9868 

98 -9912 

99 9956 


9872 
9917 
9961 


9877 
9921 
9965 


9881 
9926 
9969 


9886 
9930 
9974 


9890 

9934 
9978 


9894 
9939 
9983 


9899 
9943 
9987 


9903 
9948 
9991 


9908 

9952 
9996 


oil 
I I 
I I 


2 
2 

2 


2 3 
2 3 
2 3 


3 4 4 
3 4 4 
3 3 4 



;oo 



ANTI-LOGARITHMS. 



•00 

•01 
•02 

•03 

•04 
•05 
•06 

•07 
•08 
•09 

iO 

•II 
•12 

•13 

•14 
•15 
•16 

•17 
•18 
•19 

20 

•21 
•22 

•23 
•24 
•25 
•26 

•27 
•28 
•29 

30 

•31 
•32 

•33 

'34 
•35 
•36 

•37 
•38 
•39 

40 

•41 
•42 

'43 

•44 
•45 
•46 

•47 
•48 

•49 



XOGO 

023 

047 
072 

096 
122 
Z48 

202 
230 

259 

288 

318 

349 
380 

413 
445 

479 
514 
549 

585 

622 
660 
698 

738 
778 
820 

862 
905 
950 

1995 
2042 

2089 
2138 

2188 
2239 
2291 

2344 
2399 
2455 
2512 

2570 
2630 
2692 

2754 
2818 
2884 

2951 
3020 

3090 



Z002 

026 
050 
074 

099 
125 
151 

178 
205 

233 

262 

291 
321 
352 

384 
4x0 

449 

483 
517 
552 

589 

626 
663 
702 

742 
782 
824 

866 
9x0 

954 
2000 
2046 
2094 
2143 

2193 
2244 
2290 

2350 
2404 
2460 

2518 

2576 
2636 
2698 

2761 
2825 
2891 

2958 
3027 
3097 



X005 

028 
052 
076 

X02 
127 
153 

180 
208 
236 

265 

294 
324 
355 

387 
419 

452 
486 
521 
556 

592 

629 
667 
706 

% 

828 

871 
9x4 

959 

2004 

205 X 
2099 
2148 

2198 
2249 



230X 2307 



2355 
2410 
2466 

2523 

2582 
2642 
2704 

2767 
283X 
2897 
2965 

3034 
3105 



X007 

030 

054 
079 

X04 
X30 
156 

183 

2XX 
239 
268 

297 
327 
358 

390 
422 

455 

489 
524 
560 

596 

633 

67X 

710 

750 
791 
832 

875 
919 

9<53 

2009 

2056 
2104 
2153 

2203 

2254 



2360 

2415 
2472 

2529 

2588 
2649 
27x0 

2773 
2838 
2904 

2972 

3041 
3112 



X009 

033 

057 
o8x 

107 
132 
159 
186 
213 
242 

27X 

300 
330 
361 

393 
426 

459 

493 
528 

563 

600 

637 

675 
714 

754 
795 
837 

879 
923 
968 

20x4 

2o6x 
2109 

2x58 

2208 

2259 

2312 

2366 
242X 

2477 

253s 

2594 
2655 

27x6 

2780 
2844 

29XX 

2979 
3048 

3"9 



X0X2 

035 
059 
084 

XO9 
135 

z6x 

X89 

2X6 

345 
274 
303 

334 
365 

396 
429 
462 

496 

567 
603 

64X 
679 
7x8 

758 
799 
84X 

884 
928 
972 

20x8 

2065 
21x3 
2x63 

2213 

2265 
2317 

2371 

2427 
24S3 

2541 
2600 
266X 
2723 

2786 
2851 
29x7 

2985 

3055 
3126 



XOX4 

038 
062 
086 

1x2 

138 
164 

19X 
2x9 
247 
276 

306 

337 
368 

400 
432 
466 

500 

535 
570 

607 

644 
683 
722 

762 
803 
845 
888 
932 
977 
2023 
2070 

2XX8 

2x68 

22X8 

2270 
2323 

2377 

2432 

2489 

2547 

2606 
2667 

2729 

2793 
2858 
2924 

2992 
3062 

3133 



016 

040 

064 
089 

"4 

1^0 

107 
194 

222 

250 

279 

309 
340 
371 

403 

435 
469 

503 

538 
574 
61X 

648 
687 
726 

766 
807 
849 

892 
936 
982 

2028 

2075 
2x23 
2x73 

2223 

2275 
2328 

2382 
2438 
2495 

2553 

26x2 
2673 

2735 

2799 
2864 

2931 

2999 

3069 

3141 



8 



Z0X9 

042 
067 
09X 

X17 

143 
X69 

197 
225 

253 

282 

312 
343 
374 
406 

439 
472 

507 

542 
578 

6x4 

652 
690 
730 

770 
8xx 

854 
897 
94X 
986 

2032 

2080 
2128 
2x78 

2228 
2280 
2333 
2388 

2443 

2500 

2559 

2618 
2679 
2742 

2805 
2871 
2938 
3006 
3076 
3148 



X02X 

045 
069 
094 

XX9 
X46 
X72 

199 
227 
256 

285 

315 
346 
377 
409 

442 
476 

510 

545 
581 

6x8 

656 
694 

734 

774 
8x6 

858 

90X 

945 
99X 

2037 
2084 

2x33 
2x83 

2234 
2286 

2339 

2393 

2449 
2506 

2564 

2624 
2685 
2748 

2812 
2877 
2944 

30x3 
3083 
3155 



12 3 



1 



4. 6 6 



I I 2 
112 
I I 2 



I 2 

I 2 

I 2 

1 2 

2 2 
2 2 
2 2 

2 2 

2 2 

2 2 

2 2 

2 2 

2 2 

2 2 

2 2 

2 2 

2 2 



7 8 9 



2223 
2223 
2233 

2233 
2233 
2233 

2233 

2233 
2233 
2233 

2334 
2334 
2334 

2334 
2334 
2334 

2344 

2344 
2344 
3 3 4 4 

3344 
3 3 4 5 
3 3 4 5 

3 3 4 5 
3 4 4 5 

3 4 4 5 



222 

222 
222 

2 
2 

2 

222 
223 
223 

2 2 

2 
2 
2 

2 

2 
2 

233 
233 

3 3 3 



ANTI-LOGARITHMS. 



301 



50 


c 

3162 


1 

3170 


2 
3x77 


3 

3184 


4 

3192 


5 


6 


7 
3214 


8 


9 


1 2 


3 


4 


5 6 


7 8 9 


3199 


3206 


322X 


3228 


I I 


2 


3 


4 4 


5 6 7 


•SI 
•52 
•53 


3236 

3388 


3243 
3319 
3396 


3251 
3327 
3404 


3258 
3334 
3412 


3266 

3342 
3420 


3273 
3350 
3428 


3281 
3357 

3436 


3289 

3365 

3443 


3296 

3373 
3451 


3304 
3381 

3459 


I 2 
I 2 

I 3 


2 
2 
2 


3 
3 
3 


4 5 
4 5 
4 5 


5 6 7 
567 
667 


•54 

•55 
•56 


3467 
3548 
3^3^ 


3475 
3556 
3639 


3483 
3565 
3648 


3491 

3573 
3656 


3499 
3664 


3508 
3589 
3673 


3516 

3681 


3524 
3606 

3690 


3532 

3614 
3698 


3540 
3622 

3707 


I 2 
I 2 
I 2 


2 
2 
3 


3 

3 


4 5 

4 5 
4 5 


667 
677 

6 7 S 


■11 

•59 


3715 
3802 
3890 


3724 
381 1 
3899 


3733 
3819 
3908 


3741 
3828 

3917 


3750 

3837 
3926 


3758 
3846 

3936 


3767 
3855 
3945 


3776 
3864 

3954 


3784 
3873 
3963 


3793 
3882 

3972 


I 2 
I 2 
I 2 


3 
3 
3 


3 

4 
4 


4 5 

4 5 

5 5 


678 
678 
678 


60 


3981 


3990 


3999 


4009 


4018 


4027 


4036 


4046 


4055 


4064 


I 2 


3 


4 


5 6 


678 


•61 
•62 
•63 


4074 
4169 
4266 


4083 
4178 
4276 


4285 


4102 
4198 
4295 


4111 
4207 
4305 


4121 
4217 

4315 


4130 
4227 

4325 


4140 
4236 

4335 


4150 
4246 

4345 


4159 
4256 

4355 


I 2 
I 2 
I 2 


3 
3 

3 


4 
4 

4 


5 6 
5 6 
5 6 


789 
7 8 9 

7 8 9 


•64 

•65 
•66 


4365 
4467 

4571 


4375 
4477 
4581 


4385 
4487 
4592 


4395 
4498 
4603 


4406 
4508 
4613 


4416 

4519 
4624 


4426 
4529 
4634 


4436 
4539 
4645 


4446 

4550 
4056 


4457 
4560 
4667 


I 2 
I 2 
I 2 


3 
3 
3 


4 
4 

4 


5 6 
5 6 
5 6 


789 
789 
7 9 10 


'67 
•68 

•69 


4677 


4688 

4797 
4909 


4599 
4808 
4920 


4710 
4819 
4932 


472X 
4831 
4943 


4732 
4842 

4955 


4742 
4853 
4966 


4864 
4977 


4764 

4875 
4989 


4775 
4887 
5000 


I 2 
I 2 
I 2 


3 
3 
3 


4 
4 

5 


5 7 

6 7 
6 7 


8 9 10 
8 9 10 
8 9 10 


70 


5012 


5023 


5035 


5047 


5058 


5070 


5082 


5093 


5105 


5117 


I 2 


4 


5 


6 7 


8 9 11 


71 
•72 

•73 


5129 
5248 

5370 


5140 
5200 

5383 


5x52 
5272 
5395 


5164 
5284 
5408 


5176 
5297 
5420 


5188 
5309 
5433 


5200 
5321 
5445 


5212 

5333 
5458 


5224 

5346 
5470 


5236 
5358 
5483 


I 2 
I 2 

I 3 


4 
4 
4 


5 
5 
5 


6 7 
6 7 
6 8 


8 10 II 

9 10 II 
9 10 II 


•74 
•75 
•76 


5495 
5623 
5754 


5§?! 
5636 

5768 


5521 
5649 
5781 


5662 
5794 


5546 
5675 
5808 


5559 
5689 
5821 


5572 
5702 
5834 


5585 
5715 
5848 


5598 
5728 
5861 


5610 
5741 
5875 


I 3 
I 3 
I 3 


4 
4 
4 


5 
5 
5 


6 8 

7 8 
7 8 


9 lo 12 
9 10 12 
9 II 12 


•77 
•78 
•79 


5888 
6026 
6166 


6180 


5916 
6053 
6x94 


6067 
6209 


6223 


5957 
6095 

6237 


5970 
6109 
6252 


5984 
626^ 


5998 
6138 
628X 


6012 
6152 
6295 


I 3 
I 3 
I 3 


4 

4 
4 


5 
6 

6 


7 8 
7 8 
7 9 


10 II 12 
10 II 13 
10 II 13 


80 


6310 


6324 


6339 


6353 


6368 


6383 


6397 


64x2 


6427 


6442 


I 3 


4 


6 


7 9 


10 12 13 


♦81 

•82 

•83 


6457 
6607 
6761 


6622 
6776 


6486 
6637 
6792 


6501 

6653 
6808 


6516 

6668 
6823 


6531 
6683 
6839 


6546 
6699 
6855 


6561 
6714 
687X 


6577 
6730 
6887 


6592 

6745 
6902 


2 3 
2 3 
2 3 


5 
5 

5 


6 
6 
6 


8 9 


II 12 14 
II 12 14 
II 13 14 


•84 

•85 
•86 


6918 
7079 
7244 


6934 
7096 
7261 


6950 
7112 
7278 


6966 
7129 
7295 


6982 
7145 
73" 


6998 
7161 
7328 


7015 
7178 
7345 


7031 

7194 
7362 


7047 

721 1 

7379 


706^ 
7228 
7396 


2 3 
2 3 
2 3 


5 
5 
5 


6 

7 
7 


8 10 
8 10 
8 10 


11 13 15 

12 13 15 
12 13 15 


•87 
•88 
•89 


I'M 

7762 


7430 
7003 
7780 


7447 
7621 

7798 


7464 
7638 
7816 


7482 
7656 

7834 


7499 
7074 
7852 


7516 
7691 
7870 


7534 
7709 

7889 


7551 
7727 
7907 


7568 
7745 
7925 


2 3 

2 4 
2 4 


5 
5 
5 


7 
7 
7 


9 10 
9 " 
9 " 


12 14 16 

12 14 16 

13 14 16 


•90 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


809X 


8110 


2 4 


6 


7 


9 " 


13 15 17 


•91 
•92 

93 


8128 
8318 
851 1 


8147 
8337 
8531 


8166 
8356 
8551 


8185 

8375 
8570 


8204 

8395 
8590 


8222 


8241 

8433 
8630 


8260 

8453 
8650 


8279 


8299 
8492 
8690 


2 4 
2 4 
2 4 


6 
6 
6 


8 
8 
8 


9 II 
10 12 
10 12 


13 15 17 

14 15 17 
14 16 18 


•94 
•95 
•96 


8710 

8913 
9120 


8730 

8933 
9141 


8750 

8954 
9162 


8770 
8974 
9183 


8790 

8995 
9204 


8810 
9016 
9226 


8831 
9036 
9247 


88SX 

9057 
9268 


8872 
9078 
9290 


8892 
9099 
93" 


2 4 
2 4 

2 4 


6 
6 
6 


8 
8 
8 


10 12 

10 12 

11 13 


14 16 18 

15 17 19 
15 17 19 


.97 
•98 
•99 


9333 
9550 
9772 


9354 
9572 
9795 


937<5 
9594 
9817 


9397 
9616 
9840 


9419 
9638 
9863 


9441 
9661 

9886 


9462 
9683 
9908 


9484 
9705 
993X 


9506 
9727 
9954 


9528 
9750 
9977 


2 4 
2 4 

2 5 


7 
7 
7 


9 
9 
9 


II 13 

II 13 
II 14 


15 17 20 

16 18 20 
16 18 20 



302 








NATURAL SINES. 














o 


X 
2 

3 


O' 


6' 


12 


18 


24' 


30 


36 


42 


48 


54' 


1' 


2' 

6 

6 
6 
6 


3' 

9 

9 
9 
9 


4' 

12 
12 

12 
12 


6' 

15 

15 
15 
15 


•0000 


0017 


0035 


0052 


0070 


0087 


0105 


0122 


0140 


0157 


3 


•0175 

•0349 
•0523 


0192 
0366 
0541 


0209 
0384 
0558 


0227 
0401 
0576 


0244 
0419 
0593 


0262 
0436 
0610 


0279 

0454 
0628 


0297 
0471 
0645 


0314 
0488 
0663 


0332 
0506 
06S0 


3 
3 
3 


4 
5 


•0698 
•0872 


0715 
0889 


0732 
0906 


0750 
0924 


0767 
0941 


0785 
0958 


0802 
0976 


0819 
0993 


0837 

lOII 


0854 
1028 


3 
3 


6 
6 


9 
9 


12 
12 


14 
14 


6 

7 
8 

9 

10 

II 

12 

13 


•1045 
•1219 
•1392 


1063 
1236 
1409 


loSo 

1253 
1426 


1097 
1271 
1444 


"15 

1288 
146 1 


1132 

1305 
1478 


1 149 
1323 
1495 


1167 
1340 
1513 


1 184 

1357 
1530 


1201 

1374 
1547 


3 
3 
3 


6 
6 
6 


9 
9 

9 


12 
12 
12 


14 
14 
14 


•1564 
•1736 


1582 

1754 


1599 
1771 


1616 

1788 


1633 

1805 


1650 
1822 


1668 
1840 


1685 
1857 


1702 
1874 


1719 
1891 


3 
3 


6 
6 

6 
6 
6 


9 

9 

9 
9 
8 


II 
II 

II 

II 
II 


14 
14 

14 
14 
14 


•I 90S 
•2079 
•2250 


1925 
2096 
2267 


1942 
2113 
2284 


1959 
2130 
2300 


1977 
2147 

2317 


1994 
2164 
2334 


201 1 
2181 
2351 


2028 
2198 
2368 


2045 
2215 
2385 


2062 
2233 
2402 


3 
3 
3 


14 
15 


•2419 

•2588 


2436 
2605 


2453 
2622 


2470 
2639 


2487 
2656 


2504 
2672 


2521 
2689 


2538 
2706 


2554 
2723 


2571 
2740 


3 

3 


6 
6 


8 
8 


II 
II 


14 
14 


i6 

17 
i8 

19 

20 

2Z 
22 
23 


•2756 
•2924 
•3090 


2773 
2940 

3107 


2790 

2957 
3123 


2807 
2974 
3140 


2823 
2990 
3156 


2840 
3007 
3173 


2857 
3024 
3190 


2874 
3040 
3206 


2890 

3057 
3223 


2907 
3074 
3239 


3 
3 

3 

3 
3 


6 
6 
6 

5 

5 


8 
8 
8 

8 
8 


II 
II 
II 

II 

II 

II 
II 
II 


14 
14 
14 

14 
14 

14 
13 
13 


•3256 
•3420 


3272 
3437 


3289 
3453 


3305 
3469 


3322 
3486 


3338 
3502 


3355 
3518 


3371 
3535 


3387 
3551 


3404 
3567 


•3584 
•3746 
•3907 


3600 
3762 
3923 


3616 
3778 
3939 


3633 
3795 
3955 


3649 
3811 

3971 


3665 
3827 

3987 


3681 

3843 
4003 


3697 
3859 
4019 


3714 
3875 
4035 


3730 
3891 
4051 


3 
3 
3 


5 
5 
5 


8 
8 
8 


24 
25 


•4067 
•4226 


4083 
4242 


4099 
4258 


4115 
4274 


4131 

4289 


4147 
4305 


4163 
4321 


4179 
4337 


4195 
4352 


4210 
4368 


3 
3 


5 
5 


8 
8 


II 
II 


13 
13 


26 
27 
28 

29 

30 
31 

32 

33 


•4384 
•4540 
•4695 


4399 

4555 
4710 


4415 
4571 
4726 


4431 
4586 

4741 


4446 
4602 
4756 


4462 
4617 
4772 


4478 
4633 
4787 


4493 
4648 
4802 


4509 
4664 
4818 


4524 
4679 

4833 


3 
3 
3 


5 

5 
5 

5 
5 

5 
5 
5 


8 
8 
8 

8 
8 

7 

7 
7 


10 
10 
10 

10 
10 

10 

10 

10 


13 

13 
13 

13 
13 

12 
12 
12 


•4848 
•5000 


4863 
5015 


4879 
5030 


4894 
5045 


4909 
5060 


4924 
5075 


4939 
5090 


4955 
5105 


4970 
5120 


4985 
5135 


3 
3 


•5150 

•5299 
•5446 


5165 

5314 
5461 


5180 

5329 
5476 


5195 

5344 
5490 


5210 

5358 
5505 


5225 

5373 
5519 


5240 
5388 
5534 


5255 
5402 

5548 


5270 
5417 
5563 


5284 
5432 
5577 


2 

2 
2 


34 
35 


•5592 
•5736 


5606 
5750 


5621 
5764 


5635 
5779 


5650 
5793 


5664 
5807 


5678 
5821 


5693 
5835 


5707 
5850 


5721 
5864 


2 

2 


5 

5 


7 
7 


10 
9 


12 
12 


36 
37 
38 

39 

40 

41 
42 

43 


•5878 
•6018 
•6157 


5892 
6032 
6170 


5906 
6046 
6184 


5920 
6060 
6198 


5934 
6074 
621 1 


5948 
6088 
6225 


5962 
6101 
6239 


5976 
6115 
6252 


5990 
6129 
6266 


6004 

6143 
6280 


2 
2 

2 


5 
5 
5 


7 
7 
7 

7 
7 

7 
6 
6 


9 
9 
9 

9 
9 

9 
9 

8 


12 
12 
II 

II 
II 

II 

I J 
11 


•6293 

•6428 


6307 
6441 


6320 

6455 


6334 
6468 


6347 
6481 


6361 
6494 


6374 
6508 


6388 
6521 


6401 
6534 


6414 
6547 


2 
2 

2 
2 
2 


4 
4 

4 
4 
4 


•6561 
•6691 
•6820 


6704 
6833 


6587 
6717 
6845 


6600 
6730 

6858 


6613 

6743 
6871 


6626 
6756 
6884 


6639 
6769 

689b 


6652 
6782 
6909 


6665 

6794 
6921 


6678 
6807 
6934 


44 


•6947 


6959 


6972 


6984 


6997 


7009 


7022 


7034 7046 


7059 


2 


4 6 


8 loj 











NATURAL SINES. 












303 




|6 


12 


18 


24 


30 


36 


42' 


48 


54 1' 


2' 


3' 


4' 5' 

1 


o 

45 


•7071 


7083 


7096 


7108 


7120 


7133 


7145 


7157 


7169 


7181 


2 


4 


6 


8 


.0 


46 

47 
48 

49 
50 

SI 

52 

53 


•7193 
•7314 
•7431 


7206 
7325 
7443 


7218 
7337 
7455 


7230 

7349 
7466 


7242 
7361 
7478 


7254 
7373 
7490 


7266 

7385 
7501 


7278 
7396 
7513 


7290 
7408 
7524 


7302 
7420 
7536 


2 
2 
2 


4 
4 
4 


6 
6 
6 


8 
8 
8 


10 
10 
10 


•7547 
•7660 


7559 
7672 


7570 
7683 


7581 
7694 


7593 
7705 


7604 
7716 


7615 

7727 


7627 
7738 


7638 
7749 


7649 
7760 


2 

2 


4 
4 


6 
6 


8 

7 


9 

9 


•7771 
•7880 
•7986 


7782 
7891 
7997 


7793 
7902 
8007 


7804 
7912 
8018 


7815 

7923 
8028 


7826 

7934 
8039 


7837 
7944 
8049 


7848 

7955 
8059 


7859 
7965 
8070 


7869 
7976 
8080 


2 
2 
2 


4 
4 
3 


5 
5 
5 


7 

7 

7 


9 
9 
9 


54 

55 


•8090 
•8192 


8100 
8202 


8111 
8211 


8121 
8221 


8131 
8231 


8141 
8241 


8151 
8251 


8161 
8261 


8171 
8271 


8181 
8281 


2 
2 


3 
3 


5 

5 


7 
7 


8 
8 


56 

57 
58 

59 
60 

61 
62 

63 


•8290 

•8387 
•8480 


8300 
8396 
8490 


8310 
8406 
8499 


8320 

8415 
8508 


8329 
8425 
8517 


8339 

8434 
8526 


8348 

8443 
8536 


8358 
8453 
8545 


8368 
8462 
8554 


8377 
8471 
8563 


2 
2 
2 


3 
3 
3 


5 
5 

5 


6 
6 
6 


8 
8 
8 


•8572 
•8660 


8581 
8669 


8590 
8678 


8599 

8686 


8607 
8695 


8616 
8704 


8625 
8712 


8634 

8721 


8643 
8729 


8652 
8738 




3 
3 


4 
4 


6 
6 


7 

7 


•S746 
•8829 
•8910 


8755 
8838 
8918 


8763 
8846 
8926 


8771 
8854 
8934 


8780 
8862 
8942 


8788 
8870 
8949 


8796 
8878 
8957 


8805 
8886 
8965 


8813 
8894 
8973 


8821 
8902 
8980 




3 
3 
3 


4 
4 
4 


6 

5 
5 


7 
7 
6 


65 


•8988 
•9063 


8996 
9070 


9003 
9078 


9011 

9085 


9018 
9092 


9026 
9100 


9033 
9107 


9041 
9114 


9048 
9121 


9056 
9128 




3 
2 


4 
4 


5 

5 


6 
6 


66 

67 
68 

69 
70 

71 
72 

73 


•9135 
•9205 
•9272 


9143 
9212 
9278 


9150 
9219 

9285 


9157 
9225 
9291 


9164 
9232 
9298 


9171 
9239 
9304 


9178 
9245 
93" 


9184 
9252 
9317 


9191 
9259 
9323 


9198 
9265 
9330 




2 
2 
2 


3 
3 
3 


5 
4 
4 


6 
6 

5 


•9336 
•9397 


9342 
9403 


9348 
9409 


9354 
9415 


9361 
9421 


9367 
9426 


9373 
9432 


9379 
9438 


9385 
9444 


9391 

9449 




2 
2 


3 
3 


4 
4 


5 
5 


•9455 

•95" 

•9563 


9461 
9516 
9568 


9466 
9521 
9573 


9472 
9527 
9578 


9478 
9532 
9583 


9483 
9537 
9588 


9489 
9542 
9593 


9494 
9548 
9598 


9500 

9553 
9603 


9505 
9558 
9608 




2 
2 
2 


3 
3 
2 


4 
4 
3 


5 
4 
4 


74 
75 


•9613 
•9659 


9617 
9664 


9622 
9668 


9627 
9673 


9632 
9677 


9636 
9681 


9641 
9686 


9646 
9690 


9650 
9694 


9655 
9699 




2 

I 


2 
2 


3 
3 


4 
4 


76 
77 
78 


•9703 
•9744 
•9781 


9707 
9748 

9785 


9711 

9751 
9789 


9715 
9755 
9792 


9720 

9759 
9796 


9724 
9763 
9799 


9728 
9767 
9803 


9732 
9770 
9806 


9736 

9774 
9810 


9740 
9778 
9813 






2 
2 
2 


1 


3 
2 


3 
3 
3 


79 

So 


•9816 
•9848 


9820 
9851 


9823 
9854 


9826 
9857 


9829 
9860 


9833 
9863 


9836 

9866 


9839 
9869 


9842 
9871 


9845 
9874 


I 





2 

I 


2 
2 


3 
2 


81 
82 

83 


•9877 
•9903 

•9925 


9880 

9905 
9928 


9882 
9907 
9930 


9885 
9910 
9932 


9888 
9912 
9934 


9890 
9914 
9936 


9893 
9917 
9938 


9895 
9919 
9940 


9898 
9921 
9942 


9900 
9923 
9943 











2 
2 
I 


2 
2 

2 


84 
85 


•9945 
•9962 


9947 
9963 


9949 
9965 


9951 
9966 


9952 
9968 


9954 
9969 


9956 
9971 


9957 
9972 


9959 
9973 


9960 
9974 












I 
I 


I 
I 


86 

87 
88 

89 


•9976 
•9986 
•9994 


9977 
9987 
9995 


9978 
9988 
9995 


9979 
9989 
9996 


9980 
9990 
9996 


9981 
9990 
9997 


9982 
9991 
9997 


9983 
9992 

9997 


9984 

9993 
9998 


9985 
9993 
9998 








I 


I 


I 


•9998 


9999 


9999 


9999 


9999 


i-ooo 


I'OOO 


I'OOO 




I'OOO 


I'oooj 

1 











304 








NATURAL 


COSINES. 




SUBTRACT. 




o 


I 
2 

3 

4 
5 
6 

7 
8 

9 

10 

II 

12 

13 

14 
15 

i6 

H 
i8 

19 

20 

21 
22 
23 
24 
25 
26 

27 
28 

29 

30 

31 
32 
33 

34 
35 

36 
37 
38 

39 
40 

41 
42 
43 

44 


O' 


6 


12 


18 


24' 


30 


36' 


42 


48' 


54 


r 


2' 


3' 


4' 


'1 


I'OOOO 


rooo 


I'CXDO 


I'OOO 


i-ooo 


I'OOO 


9999 


9999 


9999 


9999 


















I 

I 
I 

I 
2 
2 




•9998 
•9994 
•9986 

•9976 
•9962 

•9945 
•9925 
•9903 


9998 
9993 
9985 

9974 
9960 

9943 
9923 
9900 


9998 

9993 
9984 

9973 
9959 

9942 
9921 
9898 


9997 
9992 
9983 
9972 
9957 

9940 
9919 
9895 


9997 
9991 
9982 

9971 
9956 

9938 
9917 
9893 


9997 
9990 
9981 

9969 
9954 

9936 
9914 
9890 


9996 
9990 
9980 

9968 
9952 

9934 
9912 
9888 


9996 
9989 

9979 

9966 

9951 

9932 
9910 

9885 


9995 
9988 

9978 

9965 
9949 

9930 
9907 
9882 


9995 
9987 
9977 

9963 
9947 

9928 

9905 
9880 


I 

I 
I 

2 
2 
2 


•9877 
•9848 


9874 
9845 


9871 
9842 


9869 
9839 


9866 
9836 


9863 
9833 


9860 
9829 


9857 
9826 


9854 
9823 


9851 

9820 




I 




I 

2 


2 

2 


2 

3 


•9816 
•9781 

•9744 
•9703 
•9659 

•9613 
•9563 
•95" 


9813 
9778 
9740 

9699 
9655 
9608 
9558 
9505 


9810 
9774 
9736 

9694 
9650 

9603 

9553 
9500 


9806 
9770 
9732 

9690 
9646 

9598 
9548 
9494 


9803 

9767 
9728 

9686 
9641 

9593 
9542 
9489 


9799 
9763 
9724 

9681 
9636 

9588 
9537 
9483 


9796 

9759 
9720 

9677 
9632 

9583 
9532 
9478 


9792 
9755 
9715 

9673 
9627 

9578 

9527 
9472 


9789 

9751 
9711 

9668 
9622 

9573 
9521 
9466 


9785 
9748 
9707 

9664 
9617 

9568 
9516 
9461 




2 

2 

2 
2 


2 
2 
2 

2 
2 

2 
3 
3 


2 
3 
3 

3 
3 

3 

4 
4 


3 
3 
3 

4 
4 

4 
4 
5 


•9455 
•9397 


9449 
9391 


9444 
9385 


9438 
9379 


9432 
9373 


9426 
9367 


9421 
9361 


9415 

9354 


9409 
9348 


9403 
9342 




2 

2 


3 

3 


4 
4 


5 

5 


•9336 
•9272 
•9205 

•9135 
•9063 

•8988 
•8910 
•8829 


9330 
9265 
9198 

9128 
9056 

8980 
8902 
8821 


9323 
9259 
9191 

9121 

9048 

8973 
8894 

8813 


9317 
9252 
9184 

9114 
9041 

8965 
8886 
8805 


931 1 
9245 
9178 

9107 
9033 

8957 
8878 

8796 


9304 
9239 
9171 

91CX) 
9026 

8949 
8870 
8788 


9298 
9232 
9164 

9092 
9018 

8942 
8862 
8780 


9291 
9225 
9157 

9085 
9011 

8934 
8854 
8771 


9285 
9219 

9150 

9078 

9003 

8926 
8846 
8763 


9278 
9212 

9143 

9070 
8996 

8918 
8838 

8755 


I 


2 
2 
2 

2 
3 

3 
3 
3 


3 
3 
3 

4 
4 

4 
4 
4 


4 
4 
5 

5 
5 

5 
5 
6 


5 
6 
6 

6 
6 

6 

7 
7 


•8746 
•8660 


8738 
8652 


8729 
8643 


8721 
8634 


8712 
8625 


8704 
8616 


8695 
8607 


8686 
8599 


8678 
8590 


8669 
8581 





3 


4 
4 


6 
6 


7 

7 


•8572 
•8480 

•8387 

•8290 
•8192 

•8090 
•7986 
•7880 


8563 
8471 
8377 

8281 
8181 

8080 
7976 
7869 


8554 
8462 
8368 

8271 
8171 

8070 
7965 
7859 


8545 
8453 
8358 

8261 
8161 

8059 

7955 
7848 


8536 
8443 
8348 

8251 
8151 

8049 
7944 
7837 


8526 
8434 
8339 

8241 
8141 

8039 

7934 
7826 


8517 
8425 
8329 

8231 
8131 
8028 
7923 
7815 


8508 

8415 
8320 

8221 
8121 

8018 
7912 
7804 


8499 
8406 
8310 

8211 
8111 

8007 
7902 
7793 


8490 
8396 
8300 

8202 
8ioo 

7997 
7891 
7782 


2 
2 
2 

2 
2 

2 
2 
2 


3 
3 
3 

3 
3 

3 
4 
4 


5 
5 
5 

5 

5 

5 
5 
5 


6 
6 
6 


8 
8 
8 

8 
8 

9 
9 
9 


•7771 
•7660 


7760 
7649 


7749 
7638 


7738 
7627 


7727 
7615 


7716 
7604 


7705 
7593 


7694 
7581 


7683 
7570 


7672 
7559 


2 
2 

2 
2 
2 

2 


4 
4 

4 

4 
4 

4 


6 
6 

6 
6 
6 

6 


7 
8 

8 
8 
8 

8 


9 
9 

10 
10 
10 

10 


•7547 
•7431 
•7314 

•7193 


7536 
7420 
7302 

7181 


7524 
7408 
7290 

7169 


7513 
7396 
7278 

7157 


7501 

7385 
7266 

7145 


7490 
7373 
7254 

7133 


7478 
7361 
7242 

7120 


7466 
7349 
7230 

7108 


7455 
7337 
7218 

7096 


7443 
7325 
7206 

7083 



The black type indicates that the integer changes. 









1 


NATURAL 


COSINES. 




SUBTRACT. ■ 


^05 




O' 


6' 


12 


18 


24' 


30 


36 


42' 


48 


54 


1' 


2' 


3' 


4' 


5' 




45 


•7071 7059 


7046 


7034 


7022 


7009 


6997 


,6984 


6972 


6959 


2 


4 


6 


8 


10 


46 
48 

49 
SO 

51 
52 
53 


•6947 
•6820 
•6691 


6934 
6807 
6678 


6921 
6794 
6665 


6909 

6782 
6652 


6896 
6769 
6639 


6884 
6756 
6626 


6871 

6743 
6613 


6858 
6730 
6600 


6845 
6717 
6587 


6833 
6704 

6574 


2 
2 
2 


4 
4 
4 


6 
6 

7 


8 
9 
9 


II 
II 
II 

II 
II 


•6561 
•6428 


6547 
6414 


6534 
6401 


6521 

6388 


650S 
6374 


6494 
6361 


6481 
6347 


6468 
6334 


6455 
6320 


6441 
6307 


2 
2 


4 

4 


7 
7 


9 
9 


•6293 
•6157 
•6018 


6280 

6143 
6004 


6266 
6129 
5990 


6252 
6II5 

5976 


6239 
6101 
5962 


6225 

6088 
5948 


621 1 

6074 
5934 


6198 
6060 
5920 


6184 
6046 
5906 


6170 
6032 
5892 


2 
2 
2 


5 

5 
5 


7 
7 
7 


9 
9 
9 


II 

12 
12 


54 
55 


•5878 
•5736 


5864 
5721 


5850 
5707 


5835 
5693 


5821 
5678 


5807 
5664 


5793 
5650 


5779 
5635 


5764 
5621 


5750 
5606 


2 
2 


5 
5 


7 

7 


9 

10 


12 
lit 


56 
58 

6o 
6i 

62 

63 


•5592 
•5446 
•5299 


5577 
5432 

52S4 


5563 
5417 
5270 


5548 

5402 
5255 


5534 
5388 
5240 


5519 
5573 
5225 


5505 
5358 
5210 


5490 
5344 
5195 


5476 
5329 
5180 


5461 

5314 
5165 


2 
2 
2 


5 
5 
5 


7 
7 
7 


10 
10 
10 


12 
12 
12 


•5150 
•5000 


5135 
4985 


5120 

4970 


5105 

4955 


5090 
4939 


5075 
4924 


5060 
4909 


5045 
4894 


5030 
4879 


5015 
4863 


3 
3 


5 
5 


8 
8 


10 

10 


13 
13 


•4848 

•4695 
•4540 


4833 
4679 

4524 


4818 
4664 
4509 


4802 
4648 
4493 


4787 
4633 
4478 


4772 
4617 
4462 


4756 
4602 
4446 


4741 
4586 

4431 


4726 
4571 
4415 


4710 
4555 
4399 


3 
3 

3 


5 
5 
5 


8 
8 
8 


10 
10 
10 


13 
13 
13 


?4 
65 


•4384 
•4226 


4368 
4210 


4352 
4195 


4337 
4179 


4321 
4163 


4305 
4147 


4289 
4131 


4274 
4115 


4258 
4099 


4242 
4083 


1 

3 


5 
5 


8 
8 


II 
II 


13 
13 


66 

67 
68 

69 
70 

71 
72 

73 


•4067 
•3907 
•3746 


4051 
3891 
3730 


4035 
3875 
3714 


4019 
3859 
3697 


4003 

3843 
3681 


3987 
3827 
3665 


3971 
381 1 

3649 


3955 
3795 
3633 


3939 
3778 
3616 


3923 
3762 
3600 


3 
3 
3 


5 

5 

5 


8 

8 
8 


II 

II 
II 


13 
13 
14 


•3584 
•3420 


3567 
3404 


3551 
3387 


3535 
3371 


3518 
3355 


3502 
3338 


3486 
3322 


3469 
3305 


3453 
3289 


3437 
3272 


3 
3 


5 
5 


8 
8 


II 
II 


14 
14 


•3256 
•3090 
•2924 


3239 
3074 
2907 


3223 
3057 
2890 


3206 
3040 
2874 


3190 
3024 
2857 


3173 

3007 
2840 


3156 
2990 
2823 


3140 
2974 
2807 


3123 
2957 
2790 


3107 
2940 

2773 


3 
3 
3 


6 
6 
6 


8 
8 
8 


II 
II 
II 


14 
14 
14 


74 
75 


•2756 
•2588 


2740 
2571 


2723 
2554 


2706 
2538 


2689 
2521 


2672 
2504 


2656 
2487 


2639 

2470 


2622 
2453 


2605 
2436 


3 


6 
6 


8 

8 


II 
II 


14 
14 


76 
77 
78 

79 
80 

8. 

82 
83 


•2419 

•2250 
•2079 


2402 
2233 
2062 


2385 
2215 
2045 


2368 
2198 
2028 


2351 
2181 
201 1 


2334 
2164 
1994 


2317 
2147 
1977 


2300 
2130 
1959 


2284 
2113 
1942 


2267 
2096 
1925 


3 

3 
3 


6 
6 
6 


8 
9 
9 


II 
II 
II 


14 
14 
14 


•1908 
•1736 


1891 
1719 


1874 
1702 


1857 
1685 


1840 
1668 


1822 
1650 


1805 
1633 


1788 
1616 


1771 

1599 


1754 
1582 


3 
3 


6 
6 


9 
9 


II 
II 


14 
14 


•1564 
•1392 
•1219 


1547 
1374 
1201 


1530 
1357 
1 184 


1513 
1340 
1167 


1495 
1323 
1 149 


1478 

1305 
1132 


146 1 

1288 

i"5 


1444 
1271 
1097 


1426 

1253 
1080 


1409 
1236 
1063 


3 
3 
3 


6 
6 
6 


9 
9 
9 


12 
12 
12 


14 
14 
14 


84 
85 


•1045 
•0872 


1028 
0854 


lOII 

0S37 


0993 
0819 


0976 
0S02 


0958 
0785 


0941 
0767 


0924 
0750 


0906 
0732 


0889 
0715 


3 
3 


6 
6 


9 
9 


12 
12 


14 
H 


86 

87 
88 

89 


•0698 
•0523 
•0349 


0680 
0506 
0332 


0663 
0488 
0314 


0645 
0471 
0297 


0628 

0454 
0279 


0610 
0436 
0262 


0593 
0419 
0244 


0576 
0401 
0227 


0558 
0384 
0209 


0541 
0366 
0192 


3 

3 
3 


6 
6 

6 


9 
9 
9 


12 

12 
12 


15 

15 
15 


•0175 


0157 


0140 


0122 


0105 


0087 


0070 


0052 


0035 


0017 


3 


6 


9 


12 


15 



A. M. 



20 



306 






NATURAL TANGENTS. 












o 


I 
2 

3 

4 
5 
6 

7 
8 

9 

10 

ZI 
12 

13 

14 
15 

i6 
17 
18 

19 
20 

21 
22 
23 
24 
25 
26 

27 
28 

29 
30 

31 

32 
33 

34 
35 

36 
37 
38 

39 
40 

41 
42 

43 
44 


O' 


6 


12 


18 

0052 


24 

0070 


30 


36 


42 


48 


64 


V 

n 


2' 

6 


3' 
9 


4' 
12 


5' 

15 


o-oooo 


0017 


0035 


0087 


0105 


0122 


0140 


0157 


0-0175 
0-0349 
0-0524 

0-0699 
0-0875 

0-1051 
0-1228 
0-1405 


0192 
0367 
0542 

0717 
0892 

1069 
1246 
1423 


0209 
0384 
0559 

0734 
0910 

1086 
1263 
144 1 


0227 
0402 
0577 

0752 
0928 

1 104 
1281 
1459 


0244 
0419 
0594 
0769 
0945 

1122 
1299 
1477 


0262 

0437 
0612 

0787 
0963 

1139 
1317 

1495 


0279 

0454 
0O29 

0805 
09S1 

1157 

1334 
1512 


0297 
0472 
0647 

0822 
0998 

1175 
1352 
1530 


0314 
0489 
0664 

0840 
1016 

1 192 
1370 
1548 


0332 
0507 
0682 

0857 
1033 
1210 
1388 
1566 




3 
3 

3 

3 
3 
3 


6 
6 
6 

6 
6 

6 
6 
6 


9 
9 
9 

9 
9 

9 
9 
9 


12 
12 
12 

12 
12 

12 
12 
12 


15 
15 
15 

15 
15 

15 
15 
15 


0-1584 
0-1763 


1602 
1781 


1620 
1799 


1638 
1817 


1655 
1835 


1673 
1853 


1691 
1871 


1709 
1890 


1727 
1908 


1745 
1926 


3 

3 


6 
6 


9 
9 


12 
12 


15 
15 


0-I944 
0-2126 
0-2309 

0-2493 
0-2679 

0-2867 

0-3057 
0-3249 


1962 
2144 
2327 

2512 

2698 

2886 
3076 
3269 


1980 
2162 
2345 

2530 
2717 

2905 
3096 
3288 


1998 
2180 
2364 

2549 
2736 

2924 

3115 
3307 


2016 
2199 

2382 

2568 
2754 

2943 
3134 
3327 


2035 
2217 
2401 

2586 

2773 

2962 

3153 
3346 


2053 
2235 
2419 

2605 
2792 

2981 
3172 
3365 


2071 
2254 
2438 

2623 
2811 

3000 
3191 
3385 


2089 
2272 
2456 

2642 
2830 

3019 
321 1 

3404 


2107 
2290 
2475 

2661 
2849 

3038 
3230 
3424 


3 
3 
3 

3 



3 

3 

'^ 



6 
6 
6 

6 

6 

6 
6 
6 


9 
9 
9 

9 
9 

9 
10 
10 


12 
12 
12 

12 
13 

13 
13 
13 


15 
15 
15 

16 
16 

16 
16 
16 


0-3443 
0-3640 


3463 
3659 


3482 
3679 


3502 
3699 


3522 
3719 


3541 
3739 


3561 
3759 


3581 
3779 


3600 
3799 


3620 
3819 


3 




10 
10 


13 
13 


16 
17 


0-3839 
0-4040 
0-4245 

0-4452 
0-4663 

0-4877 
0-5095 
0-5317 


3859 
4061 
4265 

4473 
4684 

4899 

5117 

5340 


3879 
4081 

42S6 

4494 
4706 

4921 
5139 
5362 


3899 
4101 

4307 

4515 
4727 

4942 
5161 
5384 


3919 
4122 

4327 

4536 
4748 

4964 
5184 
5407 


3939 
4142 

4348 

4557 
4770 

4986 
5206 
5430 


3959 
4163 
4369 

4578 
4791 

5008 
5228 
5452 


3979 
4183 
4390 

4599 
4813 

5029 
5250 
5475 


4000 
4204 
441 1 

4621 
4834 

5051 
5272 

5498 


4020 
4224 
4431 

4642 
4856 

5073 
5295 
5520 


3 
3 
:> 

4 
4 

4 
4 

4 


8 


10 
10 
10 

II 
II 

II 
II 
II 


13 
14 
14 

14 
14 

15 
15 
15 


17 
17 
17 

18 
18 

18 

18 
19 


0-5543 
0-5774 


5566 
5797 


5589 
5820 


5612 

5844 


5635 
5867 


5658 
5890 


5681 
5914 


5704 
5938 


5727 
5961 


5750 
5985 


4 

4 


8 
8 


12 
12 


15 
16 


19 

20 


0-6009 
0-6249 
0-6494 

0-6745 
0-7002 

0-7265 
0-7536 
0-7813 


6032 
6273 
6519 

6771 
7028 

7292 

7563 

7841 


6056 
6297 
6544 

6796 
7054 

7319 
7590 
7869 


6080 
6322 
6569 

6822 
7080 

7346 
7618 
7898 


6104 
6346 
6594 

6847 
7107 

7373 
7646 
7926 


6128 

6371 
6619 

6873 
7133 

7400 
7673 
7954 


6152 

6395 
6644 

6899 
7159 

7427 
7701 
7983 


6176 

6420 
6669 

6924 
7186 

7454 
7729 
8012 


6200 
6445 
6694 

6950 
7212 

7481 

7757 
8040 


6224 
6469 
6720 

6976 

7239 

7508 

7785 
8069 


4 
4 
4 

4 
4 

5 
5 
5 


8 
8 
S 

9 
9 

9 

9 
9 


12 
12 
13 

13 
13 

14 
14 

14 


16 
16 
17 

17 
18 

18 
18 
19 


20 
20 
21 

21 
22 

23 
23 
24 


0-8098 
0-8391 


8127 
8421 


8156 
8451 


8185 
8481 


8214 
8511 


8243 
8541 


8273 
8571 


8302 
8601 


8332 
8632 


8361 
8662 


5 
5 


10 
10 


15 
15 


20 
20 

21 
21 
22 

23 


24 
25 

26 

27 
28 

29 


0-8693 
0-9004 
0-9325 

0-9657 


8724 
9036 

9358 

9691 


8754 
9067 
9391 

9725 


8785 
9099 
9424 

9759 


8816 
9131 

9457 

9793 


8S47 
9163 
9490 

9827 


8878 
9195 
9523 

9861 


8910 
9228 
9556 
9896 


8941 
9260 
9590 

9930 


8972 
9293 
9623 

9965 


5 
5 
6 

6 


10 
II 
II 

II 


16 
16 
17 

17 









































NATURAL 


TANGENTS. 










907 


o 

45 


O' 1 6' 


12 


18 


24 


30' 


36 


42 


48 


54 


1' 

6 


2' 

12 


3' 

18 


4' 

24 


5' 
30 


1*0000 


0035 


0070 


0105 


0141 


0176 


0212 


0247 


0283 


0319 


46 
48 

49 
50 

51 
52 
53 


1-0355 
1-0724 
I -1 106 


0392 
0761 
1145 


0428 

0799 
1 184 


0464 
0837 
1224 


0501 

0875 
1263 


0538 
0913 

1303 


0575 
0951 
1343 


0612 
0990 
1383 


0649 
1028 
1423 


0686 
1067 
1463 


6 
6 

7 


12 

13 
13 


18 

19 

20 


25 
25 
26 


31 
32 

33 


1-1504 
1-1918 


1544 
i960 


1585 
2002 


1626 
2045 


1667 
2088 


1708 
2131 


1750 
2174 


1792 
2218 


1833 
2261 


1875 
2305 


7 
7 


14 
14 


21 

22 


28 
29 


34 
36 


1-2349 
1-2799 
1-3270 


2393 
2846 

3319 


2437 
2892 

3367 


2482 
2938 
3416 


2527 
2985 

3465 


2572 
3032 
3514 


2617 
3079 
3564 


2662 
3127 
3613 


2708 

3175 
3663 


2753 
3222 

3713 


8 
8 
8 


15 
16 
16 


23 
24 

25 


30 
31 
33 


38 
39 
41 


54 
55 


1-3764 
I -428 1 


3814 
4335 


3865 
4388 


3916 
4442 


3968 
4496 


4019 
4550 


4071 
4605 


4124 
4659 


4176 
4715 


4229 
4770 


9 
9 


17 

18 


26 
27 


34 
36 


43 
45 


56 
57 
58 

59 
60 

61 
62 
63 


1-4826 

I '5399 
1-6003 


4882 

5458 
6066 


4938 
5517 
6128 


4994 
5577 
6191 


5051 
5637 
6255 


5108 

5697 
6319 


5166 

5757 
6383 


5224 
5818 
6447 


5282 
5880 
6512 


5340 
5941 
6577 


10 
10 
II 


19 
20 
21 


29 
30 
32 


38 
40 
43 


48 
50 
53 


1-6543 
17321 


6709 
7391 


6775 
7461 


6842 
7532 


6909 
7603 


6977 
7675 


7045 
7747 


7113 

7820 


7182 
7893 


7251 
7966 


II 
12 


23 

24 


34 
36 


45 
48 


56 
60 


1-8040 
1-8807 
1-9626 


8115 

8887 
9711 


8190 
8967 
9797 


8265 
9047 
9883 


8341 
9128 
9970 


8418 
9210 
0057 


8495 
9292 
0145 


8572 

9375 
0233 


8650 
9458 
0323 


8728 
9542 
0413 


13 
14 
15 


26 
27 
29 


38 
41 
44 


51 

55 
58 


64 

68 

73 


^4 
65 


2-0503 
2-I44S 


0594 
1543 


0686 
1642 


0778 
1742 


0872 
1842 


0965 
1943 


1060 
2045 


"55 

2148 


1251 
2251 


1348 

2355 


16 
17 


31 
34 


47 
51 


63 

68 


78 
85 


66 
67 
68 

69 
70 

71 

72 

73 


2-2460 
2-3559 
2-4751 


2566 

3673 
4876 


2673 

3789 
5002 


2781 
3906 
5129 


2889 
4023 

5257 


2998 
4142 
5386 


3109 
4262 
5517 


3220 
4383 
5649 


3332 
4504 
5782 


3445 
4627 

5916 


18 
20 
22 


37 
40 
43 


55 
60 

65 


73 
79 
87 


91 

99 
108 


2-6051 
27475 


6187 
7625 


6325 
7776 


6464 
7929 


6605 
8083 


6746 
8239 


6889 
8397 


7034 
8556 


7179 
8716 


7326 
8S7S 


24 
26 


47 

52 


71 

78 


95 
104 


119 

130 


2-9042 
3-0777 
3-2709 


9208 
0961 
2914 


9375 
1 146 
3122 


9544 
1334 
3332 


9714 
1524 
3544 


9887 
1716 

3759 


0061 

1910 
3977 


0237 

2106 
4197 


0415 

2305 
4420 


0595 

2506 
4646 


29 
32 
36 


58 

64 
72 


87 

97 

loS 


116 
129 
144 


144 
161 
180 


74 
75 


3-4874 
37321 


5105 

7583 


5339 

7848 


5576 
8118 


5816 
8391 


6059 
8667 


6305 
8947 


6554 
9232 


6806 
9520 


7062 
9812 


41 

46 


81 
93 


122 
139 


163 
186 


203 
232 


76 

77 
78 

79 
80 

81 
82 
83 

l^ 
85 


4-0108 

4-3315 
4-7046 


0408 
3662 
7453 


0713 

4015 
7867 


1022 

4373 
8288 


1335 
4737 
8716 


1653 
5107 
9152 


1976 
5483 
9594 


2303 
5864 
0045 


2635 
6252 

0504 


2972 
6646 
0970 


53 
62 

73 


107 
124 
146 


160 
186 
220 


214 
248 
293 


267 
310 
366 


5-1446 
5-671 


1929 
5730 


2422 
5-789 


2924 
5-850 


3435 
5-912 


3955 
5-976 


4486 
6-041 


5026 
6-107 


5578 
6-174 


6140 
6-243 


87 175 


263 


350 


438 


6-314 

7-115 
8-144 

9-51 
11-43 


6-386 
7-207 
8-264 

9-68 
11 -66 


6-460 
7-300 
8-386 

9-84 
11-91 


6-535 
7-396 

8-513 

1002 
12-16 


6-6i2 

7-495 
8-643 

10-20 
12-43 


6-691 
7-596 
8-777 

10-39 
12-71 


6-772 
7-700 
8-915 

10-58 
13-00 


6-855 
7-S06 
9-058 

10-78 
13-30 


6-940 
7-916 
9-205 

10-99 
13-62 


7-026 
8-028 
9-357 

11-20 
13-95 


Differences 

untrustworthy 

here. 


86 

87 
88 

89 


14-30 
19-08 
28-64 


14-67 
1974 
30-14 


15-06 

20-45 
31-82 


15-46 
21-20 

33-69 


15-89 
22-02 
35-80 


16-35 
22-90 

38-19 


16-83 
23-86 
40-92 


17-34 
24-90 

44-07 


17-89 
26-03 

47-74 


18-46 
27-27 
52-08 




57-29 


63-66 


71-62 


81-85 


95-49 


1 14-6 


143-2 


191-0 


286-5 


573-0 










J 



The black type indicates that the integer changes. 



308 



RADIANS AND DEGREES, 



o 


Rad. 





Rad. 





Rad. 


/ 


Rad. 


/ 


Rad. 




Rad. 


Degrees. 


o 


o-oooo 


30 


0-5236 


60 


1-0472 





-0000 


30 


•0087 




o-ooi 

0-002 


0*o6 
o-ii 


I 


o'Oi75 


31 


0-541 1 


61 


1-0647 


I 


-0003 


31 


•0090 




0003 


0-17 


2 


0-0349 


32 


0-5585 


62 


I -082 1 


2 


-0006 


32 


-0093 








3 


0-0524 


33 


0-5760 


63 


1-0996 


3 


•0009 


33 


-0096 




0-004 
0-005 


0*23 
0-29 


4 


0-0698 


34 


0-5935 


64 


I • 1 1 70 


4 


•0012 


34 


•0099 




0006 


0-34 


5 


0-0873 


35 


0-6109 


65 


I-I345 


5 


-0015 


35 


*0I02 








6 


0-1047 


36 


0-6283 


66 


1-1519 


6 


-0017 


36 


-0105 




0007 
0008 


0*40 
0-46 


7 


0-1222 


37 


0-6458 


67 


I -1694 


7 


-0020 


37 


*oio8 




0009 


0-52 


8 


0-1396 
O-I57I 


38 
39 


0-6632 
0-6807 


68 


1-1868 


8 


•0023 
•0026 


38 
39 


-01 1 1 








9 


69 


I -2043 


9 


•OII3 




001 


* 
0-57 
























002 


i-iS 


10 


0-1745 


40 


0*6981 


70 


1*2217 


10 


-0029 


40 


•0II6 




0-03 


1*72 


IX 


0-1920 


41 


0-7156 


71 


1-2392 


II 


-0032 


41 


*oii9 




0-04 


2*29 


12 


0-2094 


42 


0-7330 


72 


1-2566 


12 


-0035 


42 


*OI22 




005 


2-86 


13 


0-2269 


43 


0-7505 


73 


1*2741 


13 


-0038 


43 


•0125 




006 


3*44 


14 


0-2443 


44 


0-7679 


74 


1-2915 


14 


-0041 


44 


•0128 




007 


4*01 


15 


0-2618 


45 


0-7854 


75 


1-3090 


15 


-0044 


45 


-OI3I 




008 


4-58 


i6 
17 


0-2793 
0-2967 


46 

47 


0-8029 
0-8203 


76 

77 


1-3265 
1-3439 


16 

17 


•0047 
-0049 


46 
47 


•0134 
•0137 




009 


5*i6 


0*1 


5-73 


i8 


0-3142 


48 


0-8378 


78 


1-3614 


18 


-0052 


48 


-0140 




0-2 


11*46 


19 


0-3316 


49 


0-8552 


79 


1-3788 


19 


-0055 


49 


•0143 




0-3 


17*19 


20 


0-3491 


50 


0-8727 


80 


1-3963 


20 


-0058 


50 


•0145 




0-4 
0-5 


22-92 
28-65 


21 


0-3665 


51 


0-8901 


81 


1*4137 


21 


*oo6i 


51 


-0148 




0-6 


34-38 


22 


0-3840 


52 


0-9076 


82 


1-4312 


22 


-0064 


52 


-OI5I 








23 


0-4014 


53 


0-9250 


83 


1*4486 


23 


-0067 


53 


•0154 




0-7 

0-8 


40*11 

45-84 


24 


0-4189 


54 


0-9425 


?4 


I -4661 


24 


-0070 


54 


•0157 




0-9 


51-57 


26 


0-4363 
0-4538 


56 


0-9599 
0-9774 


85 
86 


1-4835 
1-5010 


26 


•0073 
-0076 


5| 
56 


'OTfin 








-0163 




I 


57-30 
























2 


114-59 


27 


0-4712 


57 


0-9948 


87 


1-5184 


27 


*oo79 


57 


*oi66 




3 


171-89 


28 


0-4887 


58 


1-0122 


88 


1-5359 


28 


•0081 


58 


-0169 








29 


0-5061 


59 


1-0297 


89 


1-5533 


29 


•0084 


59 


-0172 




4 
5 


229*18 

286*48 


30 


0-5236 


60 


1-0472 


90 


1-5708 


30 


*oo87 


60 


•0175 




6 


343*77 



WEIGHTS AND MEASURES. 



1 . Measure of Distance : or Linear Measure. 



1 2 inches = 1 foot. 
= 1 yard. 



3 feet 



17G0 yards = 1 mile. 
1 00 links = 1 chain. 
= 22 yards. 

Less important : 

1 pole or rod = 5-5 yards. 
1 furlong =220 yards. 
1 cable =G08 feet. 

1 sea mile — GOSO feet. 



10 millimetres = 1 centimetre. 
100 centimetres = 1 metre. 
1000 metres = 1 kilometre. 



1cm. =0-394 in. 
Im. =1-09 yds. 

= S9-4in. 
1 km. = 621 mi 
1 in. = 2-54 cm. 



10 decimetres = 1 metre. 



WEIGHTS AND MEASURES (continued). 



309 



2. Measure of Area : or Square Measure. 



144 sq. inches = 1 sq.foot. 
9 sq.feet = 1 sq. yard. 
484 sq. yards = 1 sq. chain. 

10 sq. chains = 1 acre. 
640 acres = 1 sq. mile. 



100 sq.cm. = 1 sq. dm. 
100 sq.dm.= 1 sq. m. 
10,000 sq. m. = 1 hectare. 
100 hects. = 1 sq. km. 



1 sq.cm. = 0*155 sq. in. 
1 hectare = 2*47 acres. 
1 sq. in. = 6 45 sq. cm. 



1 sq. rod or pole = 30 '25 sq. yarda. 
1 rood = 40 sq. poles. 

4 roods s= 1 acre. 



1 are =100 sq. metres. 



3. 



Measure of Volume : or Cubic Measure, Measure of Capacity, and 
Liquid Measure. 



1728 cu. in. = 1 cu. ft. 
27 cu. ft. = 1 cu. yd. 
1 gallon = 277 cu. in. 
1 cu. ft. = 6 24 gallons. 

2 pints = 1 quart. 
4 quarts = 1 gallon. 
8 gallons = 1 busheL 
8 bushels =1 quarter. 

4. Measure of Weight. 



1000 c. c. = 1 cu. dm. 

= 1 litre. 



1 c. c. = 00610 cu. in. 
1 litre = 0-220 gal. 
1 cu. in. = 16'4 c. a 



100 centilitres = 1 litre. 
100 litres = 1 hectolitre. 



16 ounces = 1 pound. 
112 pounds = 1 cwt. 
20 cwt. = 1 ton. 



1000 grms. 
1000 kgs. 



1kg. 

1 metric tonne. 



7000 grains = 1 pound. 
14 pounds =1 stone. 



1 kg. = 2'20 lbs. 
1 tonne = 0*984 ton. 
1 lb. =454 grms. 

28 pounds = 1 quarter, 
100 pounds =1 cental. 



CONSTANTS. 

7r = 3-1416 logio7r = 0-4971 

1 radian = 57*296 degrees. 

e = 2-7183 log,o6 = 0-4343 

log,i^= 2-3026 logioiV; \og,,N= 0-4343 logeiV^. 

Earth's mean radius = 3960 miles = 6-371 x 10' cm. 

A velocity of 60 miles per hour = 88 feet per second. 

A velocity of 1 knot = 1 sea-mile per hour = 1-7 feet per sec. (nearly), 

g = 32-2 ft. per sec. per sec. or 981 cm. per sec. per sec. 

Length of seconds pendulum (Greenwich) = 39-139 in. = 99*413 cm. 

1 atmosphere = 760 mm, or 29*9 in. of mercury = 103 kg. per sq. 
cm. = 14-7 lbs. per sq. in. 

Velocity of sound in air is about 1100 ft. per sec. = 3-3 x 10* cm. 
per sec. 

Velocity of light in vacuo = 186,300 miles per sec. = 3x10^*' cm. 
per sec. 

20—3 



ANSWERS TO EXERCISES 
EXERCISES I. 

1. 8-7 lbs. 

2. 99 lbs. 4. 72 lbs. 5. 13 lbs. 

6. Extracting component 8*66 lbs. ; bending component 6 lbs. 

7. 78 lbs. in the short wire and 26 lbs. in the long one. 

8. Pressure 98-5 lbs. ; sliding force 17*4 lbs. 

9. Between 11° and 12°. 10. -33 ton. 

11. 46-7 lbs. 12. 47 lbs. and 63 lbs. 13. 223 lbs. 
14. Resultant =41 '4 lbs. and acts along the Line bisecting L PjOP^ in the 

opposite direction from OP^. 15. 25*8 lbs. 

EXERCISES II. 

1. 16-4 lbs. 2. 1^ = 110-77 lbs. Pressure on fulcrum =614-77 lbs. 

3. 11-05 ins. from end weighted with 35 lbs. 

4. 3^ ft. from end weighted with 36 lbs. 5. 41-8 ins. 

6. 64-5 lbs. per sq. in. above atmosphere. 7. 12-5 lbs. 

8. 17-7 lbs. 9. -866 ton. 

10. 3-83 ins. nearly from the force of 10 lbs. 11. 69-2 lbs. 

12. 3-55 tons, 2-46 tons. 13. 17,000 lbs. 

14. Halfway along a line joining the apex to one-third of the base. 

15. 160 Iba. 16. Loses ^-^^ lb. per lb. 17. 3 owt. 

EXERCISES m. 

1. 1,080,000 ft. -lbs. 2. 1,752,000 ft. -lbs. 3. 127-4 h.p. 

4. 22^9tH.p. 5. 352 cub. ft. 6. 42,240,000 ft. -lbs. 

7. 27,456 ft. -lbs. 8. 68Ah.p. 9. 450 h.p. 

10. 5,430,000 ft. -lbs. 11. -29 h.p. 12. 119,000,000 ft. -lbs. 

13. 179-2 H.p. ; 224 H.P. 14. 218 h.p. 15. 2,704,000 ft. -lbs. 

16. 720 tons. 17. 247,000 ft. -lbs.; 112,700 ft. -lbs.; 134,300 ft. -lbs. 

EXERCISES IV. 

1. 195-45 lbs. 2. 99-77% efficiency. 

3. Required pull, 166f lbs.; Efficiency = 83-33% . 

Mechanical advantage = 3, Velocity ratio =3-6:1. 

4. 132-2 H.P. 

5. Necessary force parallel to plane = 1| tons wt. 
Necessary force parallel to base =lJton8 wt. 

6. 6126 lbs. 7. 12,600 in. -lbs. 8. 5714 h.p. 

9. F^=326-9; 17 = 1552 lbs; 77 = 8-5%. 10. 22-3 h.p. 

11. 7i H.p. 12. 99-7 H.P. 13. 34-3 h.p. 

14. 16-7 H.P, 15. 91%. 16. 26%. 



ANSWERS TO EXERCISES 3H 

EXERCISES V. 

1. 40-1 ft. per sec. 2. 6504 miles per hr. 3. 100-6 ft. 

4. Heights faUen, 9-82 ft. ; 0-9676 ft.; 0-0966 ft.; Average velocities, 

98-21 ft. per sec. ; 06-76 ft. per sec. ; 96-616 ft. per sec. 

5. 1-47 ft. per sec.^ 6. 80 miles per hour after -2 hour. 

7. 7-56 miles. 8. 3657 cms. per min. 

9. 6-87 ft. per sec. per sec. 10. 75 miles per hr. 11. 100 yds. 

12. 22 ft. per sec. per sec. 13. 1370 yds. 14. 60 miles per hour. 

EXERCISES VI. 

1. 65 feet per sec. at 37° to the direction of the train's motion. 

2. 43 ft. per sec. 3. 10 miles an hour from the N.W. 

4. 46-2 ft. per sec. ; 83-25° to the circumference. 5. 27-3 sees. 

6. 6 sees. 7. f^ min. ; 1;^ mins. ; 20^^ sees. 8. 35-5 ft. per sec. 
9. 8-03 miles. 10. 25° 37' N. of E.; 3 89 miles per hour. 

EXERCISES VII. 

1. 27 ft. -lbs. 2. 1880 yds. 3. 8430 ft. -lbs. 

4. 6170 lbs. 5. Energy =93,000 ft. -lbs.; Pressure exerted = 18,600 lbs. 

6. 3 lbs. 7. 220 ft. tons; -8 ft. per sec. increase in velocity. 

8. 25-6 ft. per sec. ; 7700 ft. 9. 31101b. sec. 

10. 1560 lbs.; 8-8 ft. 11. 760 feet. 12. 2880 feet. 

13. 45-6 tons. 14. 418 lbs. 
15. •64 ft. per sec.2; 38-4 ft. per sec. 16. 2904 lbs. 

EXERCISES VIII. 

1. 3880 lbs. 2. 6-24 tons. 3. 845 lbs.; 1000 lbs.; 1155 lbs. 

4. 221 lbs. 5. 12-9 lbs. 6. 68-8 ft. per sec. ; more. 

7. 128 ft. per sec. 8. 8-8 ft. per sec. ; come to rest. 

EXERCISES IX. 

1. -0006. 2. -Ift. 3. --0005. 

4. Stress 16,000 lbs. per sq. in.; strain -000625; ^ = 25,600,000. 

5. -0256 ft. = J in. nearly. 6. 12,000 in. -lbs. ; 345in..lbs. 
7. 3410 lbs. per sq. in. 8. 2-026 ins. 9. 14-02 tons. 

10. 48,600 lbs. per sq. in. 

11. Modulus of elasticity =22,600,000 lbs. persq. in. 

12. Work done =6 in. lbs. 13. 2700 ft. 14. 6 in. lbs. 
15. 6-45 tons per sq. in. 16. 3^ ins. diameter. 

17. -00074 in. longitudinal; -000185 in. transverse. 18. 2180 lbs. 

EXERCISES X. 

1. 30 tons. 2. fin., double row at 4-inch pitch; 66%. 3. -57. 

4. Resistance to shearing of rivets = 18-85 tons; resistance to tearing of 

plates = 18 tons; thickness of cover plates should be ^^in. ; eflSciency 
= 66-7%. 

5. 6 rivets; yes. 6. 120 tons. 7. 5 J ins. 9. 2-23 ins. nearly. 
10. 3-71 ins. 11. 48-6 lbs. per sq. in. about. 12. -36710. 



312 ANSWERS TO EXERCISES 



EXERCISES XL 

1. AB=7; BC=~11; CA= +15; £2)= -13 tons. 

2. AB=A'B':= +4-62; BD=B'D= -462; AD=A'D= -2-30; 
55'= +2-30 tons. 

3. Top bars, +48-4, 62*2, 41-4, 20*5 tons. 
Bottom bars, -24-6, 656, SIS, 308, 101 tons. 
Diagonals, ±48*4, ± 13*8, remainder ±20*9 tons. 

4. BC= +3-5; CD= +20; AE= -559; EF= -298; ^i)= -2-24; 

£^=+212; i^C7=-l-86; C7J'= +9*4 tons. 

5. Force in DD'= 6 -5 tons. 6. 2 69 tons. 
7. - 10 tons in stay; +8*8 tons in each leg. 

EXERCISES Xn. 

1. B.M.'s 1410, 870 and 480 lbs. ft. 
s.F.'s 180, 180 and 80 lbs. 

2. B.M.'s, 0, 1400, 2600, 3600, 4400, 5000, 5400, 5600, 5625 (centre) tons ft. 
s.F.'s, 150, 130, 110, 90, 70, 50, 30, 10, (middle). 

3. B.M. (at centre) = 62,600 lbs. ft. 
s.F. (at centre) =0. 

B.M. (16 ft. from end) =52,500 lbs. ft. 
s.F. (15 ft. from end) =2000 lbs. 

4. B.M. (at fixed end) =91,876 lbs. ft. 
B.M. (15 ft. from fixed end) =30,000 lbs. ft. 
B.M. (26 ft. from fixed end) =7500 lbs. ft. 
s.F. (at fixed end) =5250 lbs. 

s.F. (15 ft. from fixed end) =3000 lbs. 
s.F. (25 ft. from fixed end) = 1500 lbs. 

5. (At fixed end) b.m. = 105,000 lbs. ft., s.P. =6000 lbs. 
(16 ft. from fixed end) b.m. =31,875 lbs. ft., s.P. =3750 lbs. 
(25 ft. from fixed end) b.m. = 7500 lbs. ft., s.P. = 1500 lbs. 

6. (At centre) b.m. =45 tons ft., s.p. =2 tons. 

(5 ft. from end nearest which is wt.) b.m. =20 tons ft., s.F. =4 tons. 
(5 ft. from other end) b.m. =10 tons ft., s.F. =2 tons. 

7. Reaction at support A=^^- tons. 
Reaction at support jB=-*v- tons. 

At C, b.m. =44 J tons ft., s.b. =2f tons. 
At D, b.m. =73 tons ft., s.F. =2f tons. 
At Ej B.M. =70 tons ft., s.F. =4f tons. 

8. If A is one end of the axle and B, C, D points distance 4 ins. apart : 
Then b.m. at A =0; at B, C, D, etc. =20 tons ins. 

s.F. at .4=2^; at jB = 2|; from B to other end = 0. 

9. At point 6 ft. from one end b.m. =6900 lbs. ft., s.F, =930 lbs. 
At point 8 ft. from one end b.m. =20,400 lbs. ft., s.F. =390 lbs. 

10. B.M. =266,250 lbs. ft. ; s.F. =4000 lbs. 

11. 96 lbs. 12. B.M. =120ton3ft.; s.F. = 1 ton. 
13. At centre b.m. =648 lbs. ft., s.p. =0. 

At 1st quarter b.m. =486 lbs. ft., s.p. =72 lbs. 
At end b.ii. =0, s.p. = 144 lbs. 



ANSWERS TO EXERCISES 313 



EXERCISES XIII. 

1, 2 ins. from the base. 2. 2-5 ins. from the base. 

3. yV of its length from the centre. 4. 1'122 ft. from the point of contact, 
5. 2-33 ins. from the centre of the rod. 6. 24-16 ins, from bottom. 
7. -622 from centre of circle. 8. 18,900 lbs. 

9. Height = basex 1-732. 10. -831 in. from base. 

11. Centre of gravity = 3-747 ins. nearly from lower flange XY. 

12. Centre of gravity is a point '83 in. nearly from the 4-in. side and 1-33 ins. 

nearly from the 3 -in. side. 

13. 2-6 ins. 14. A point 2-75 ft. from the end weighted with 2 lbs. 

91 
15. tV ft. 16- r7i from right-hand end. 

lb 

17. 4f ins. from base. 18. 2-97 ft. 

EXERCISES XIV. 

1. 0-179. 2. 10 lbs, 3. -15. 

4. (i) 371-7 lbs, (ii) 370-7 lbs. (iii) 370 lbs. 

5. 10 tons at 30% to horizon. 6. 60-5 ft. 8. 4-6. 
9. 9 ins. 10. 8°; 800 sq. ft. 11. 91 lbs.; 28%. 12. 50 ft. 

EXERCISES XV. 

1. 2,744,000 ft. -lbs. 2. 2-57 lbs. 3. 47-5 revs, per min. 

4. 1410 lbs. 5. 28-6 revs, per min. 6. 15 miles per hr. 

7. Centrifugal force = 8 tons approx. ; 38-8 tons at 12° to vert. 

8. 42-6 miles per hr. 9. 17,000 ft, -lbs. 10. 38-2 revs, per min. 
11. 80,000 ft. 12. 6000 yds. 

EXERCISES XVI. 

1. (a) 84°; (6) '55 of stroke from the back dead point. 

2. (1) 12,403 lbs,; (2) 3100-75 lbs, 4. 640 ft, per min. 

6. 1-28: 1; 4 ins. 

EXERCISES XVII. 

1. 3 ft, 10 ins. nearly, 2. 18. 

3. Difference in tension =683'3 lbs, ; tension on sides = 291-7 lbs. and 875 lbs. 

4. 3 tons, 5. 250 lbs, 6. 675 lbs. 

7. 37-3 lbs. 8. 912 revs, per min, 9. -91 h,p, 

10. 13-37 lbs. 11. SJins. 12. 7^ ins, 13. 55-8 ft. 

14. 12 x^^ 15. 465,23-8. 16. 45-3% : 27-2 H.r. 



INDEX 



Acceleration 92-100 

relation to force 117 
Angle of friction 225 

of repose 233 
Angular velocity 243 

Back gear for lathes 287 
Balancing rotating parts 250 
Beams and girders 188-200 

bending moment 189 

reactions 21 

shearing force 189 
Belt gearing 273-279 
Bending moments 189-200 
Bevel gearing 281 
Bicycle two-speed gear 73 
Bow's notation 3 
Brake horse-power 78 

Cams 267 

Cantilevers 190-193 

Cast iron, stress-strain diagram of 

143 
Cement and concrete 144 
Centrifugal and centripetal force 

245 
Centroid and centre of gravity 203- 

220 
Clerk Maxwell diagrams 178 
Columns 156 
Compressive stress 140 
Cone, centre of gravity of 212 
Conservation of energy 40 
Counterbracing 175 
Couples 31 
Crank and connecting-rod mechanism 

258 
Crow-bar 53 

Curved path, motion in 242-256 
Cycloid curve 104 
Cylinders, strength of 169 



Deficient frames 174 
Diametral pitch 280 
Differential pulley block 



67 



Efficiency 

of machines 55 
of riveted joints 



168 



EfEort curve 42 

mean 49 
Elastic bodies 139 

limit 141 
Energy 

conservation of 40 

definition and kinds 39 

useful 41 
Equilibrant 8 
Equilibrium 8 

kinds of 221 

under three forces 25 
Experiments 

bicycle two -speed gear 7 

centre of gravity and centroid 
220 

errors of 13 

friction 238 

inclined plane 60 

moments 18 

polygon of forces 12 

reaction of jet 133 

roof-truss 184 

triangle of forces 4 

Weston pulley block 72 

wire, strength of 147 

Factor of safety 151 
Force 

polygon 11 

triangle of 3 

unit of 2 

see also effort, reaction, resistance, 
etc., and various kinds of 
machines and structures 
Framed structures 174-185 
Friction 224-240 

angle of 225 

rolling 228 

static and kinetic 224 

Gearing 273-290 
Gear trains 286 
Governor 248 
Gravity acceleration 97 
centre of 203-220 

Hodograph 242 
Hooke's law 139 



INDEX 



315 



Horse-power 
brake 78 
definition 38 
indicated 77 

Idle gear wheels 285 
Impact and impulse 129 
Inclined plane 57-63, 229-233 
Indicated horse-power 77 
Instantaneous centre 259 
Involute gearing 280 

Kinetic energy 40, 113 
friction 224 

Lathe back gear 287 
headstock gear 276 
lead screw drive 289 
Leverage 16 
Lever safety valve 23 
Limit, elastic 141 
Line of pressure 11 
Link and vector polygon construction 

27, 32 
Lubrication, see Friction 

Machines 62-81 

actual performance 70 

reversing 66 
Mean effort 49 
Mechanical advantage 53 
Mechanisms 268-270 
Method of moments or sections 183 
Module of toothed gearing 280 
Modulus, elastic 145 
Moments 

of forces 16-33 

method of, for frames 183 
Momentum 119-137 
Motor tracks 247 

Newton's laws of motion 1, 21, 117, 
124 

Parabola, centroid of 219 

Pawl and ratchet mechanism 270 

Pile driver 136 

Pillars 156 

Pipe3, strength of 169 

Pitch circle 280 

Planing machine, belt drive 278 

Poisson's ratio 146 

Polygon of forces 11 

Potential energy 40 

Power (.«ee also Horse-power) 38 

Projectiles 252-256 

Pulley tackle 67-70 

Pyramid, centre of gravity of 211 

Quadrilateral, centroid of 216 
Quick-return mechanism 263 



Rack and pinion 281 
Railway curves 247 
Ratchet mechanism 270 
Reaction 

Newton's law 124 

of beams 21 
Reciprocal figures 178 
Recoil of guns 133 
Redundant frames 175 
Relative velocity 103-111 
Repose, angle of 233 
Resilience 152 
Resistances 45 
Resultant, definitions of 3 
Reversing gear train 280, 290 
Reversing machines 66 
Rigidity modulus 145 
Ritter's method for frames 186 
Riveted joints 161-1G9 
Rolling friction 228 
Roof-truss 177-185 
Rotating bodies 75 

Safety valve 23 
Scalar quantities 1 
Screw 

jack 65 

with friction 233 

without friction 63 
Sections, method of, for frames 183 
Semicircle, centroid, etc. of 220 
Shear 

in beams 189-200 

legs 184 

modulus 145 

stress 140 
Ships, relative velocities of 108 
Space curve 85 
Speed, see Velocity 
Speed-cones 276 
Spiral gearing 282 
Stability of wall 22 
Steam-engine foundation, thrust on 

5 
Steel, stress-strain, diagram of 142 
Strain 139 
Stress 

definition 139 

dynamic 153 

-strain diagrams 141-150 

temperature 156 

working 151 
Struts 156 

Tensile stress 140 

Timber, strength of 144 

Toggle mechanism 265 

Toothed gearing 279-290 

Torque 76 

Trapezium, centroid of 21:i 

Triangle, etc., centroid of 210 



316 



INDEX 



Triangle of forces 3 
Tripods 184 

Useful energy 41 

Vector polygon construction 10 

Vector quantities 1 

Velocity 

angular 243 

curves 87 

definitions 83 

in mechanisms 259 

ratio in gearing 275, 279, 283 

ratio of machines tS 

relative 103-111 

uniform 83 



Velocity 

variable 84 
Virtual centre 2G0 

Warren girder 180 

Watt's parallel motion 262 

Weight as unit of force 1 

Weston's pulley block 68 

Work 36 

resistance, against 44 
variable force done by 42 

Working stress 151 

Worm gearing 282 



Yield point 141 
Young's modulus 



145 



camdridgk: printed by j. b. r:;ACE, m.a., at the university press. 



